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Concept Review

Differential Calculus

Basic Differentiation Rules

Differentiating from the definition works, but evaluating lim⁡h→0f(x+h)−f(x)h\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} by hand for every new function is hopeless. This topic builds the algebra of derivatives: a short list of rules, each proved from the definition, that lets you differentiate any polynomial, any rational function, and any product or quotient of functions whose derivatives you already know. Everything that comes later in the course — the chain rule, implicit differentiation, related rates, curve sketching, optimisation — assumes you can apply these rules quickly and without algebra errors.

Standing assumptions

Unless stated otherwise, ff and gg are real functions and aa is an interior point of their domains. "ff is differentiable at aa" means the limit f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} exists as a finite real number. Recall two facts used repeatedly below: differentiability at aa implies continuity at aa, and a derivative statement is always local — it says nothing at points outside the domain.

Linearity of differentiation

Theorem (constant multiple and sum rules). If ff and gg are differentiable at aa and c∈Rc\in\mathbb{R} is a constant, then cfcf and f+gf+g are differentiable at aa and (cf)′(a)=c f′(a),(f+g)′(a)=f′(a)+g′(a).(cf)'(a)=c\,f'(a),\qquad (f+g)'(a)=f'(a)+g'(a).

Derivation sketch. cf(a+h)−cf(a)h=c⋅f(a+h)−f(a)h\dfrac{cf(a+h)-cf(a)}{h}=c\cdot\dfrac{f(a+h)-f(a)}{h} and (f+g)(a+h)−(f+g)(a)h=f(a+h)−f(a)h+g(a+h)−g(a)h\dfrac{(f+g)(a+h)-(f+g)(a)}{h}=\dfrac{f(a+h)-f(a)}{h}+\dfrac{g(a+h)-g(a)}{h}; now apply the limit laws for constant multiples and sums, which are legitimate exactly because both limits are assumed to exist.

Combining the two and using f−g=f+(−1)gf-g=f+(-1)g gives the general form: for constants c1,…,cnc_1,\dots,c_n and functions differentiable at aa, (∑k=1nckfk)′(a)=∑k=1nckfk′(a).\left(\sum_{k=1}^{n}c_k f_k\right)'(a)=\sum_{k=1}^{n}c_k f_k'(a).

Caution. Linearity covers sums and constant multiples only. There is no rule (fg)′=f′g′(fg)'=f'g' and no rule (f/g)′=f′/g′(f/g)'=f'/g'; both are false (see Common mistakes).

The power rule

Theorem (power rule). For a fixed exponent rr, ddxxr=r xr−1\dfrac{d}{dx}x^{r}=r\,x^{r-1}, at every xx for which both xrx^{r} and xr−1x^{r-1} make sense. The precise statement depends on the kind of exponent:

Exponent rrWhere xrx^r is definedWhere ddxxr=rxr−1\frac{d}{dx}x^r=rx^{r-1} holdsHow it is proved
r=nr=n, positive integerall x∈Rx\in\mathbb{R}all x∈Rx\in\mathbb{R}binomial theorem, or the factorisation bn−an=(b−a)(bn−1+bn−2a+⋯+an−1)b^n-a^n=(b-a)(b^{n-1}+b^{n-2}a+\cdots+a^{n-1})
r=0r=0x≠0x\ne 0 (and x0=1x^0=1)all x≠0x\ne 0; ddx(1)=0\frac{d}{dx}(1)=0constant function
r=−mr=-m, mm positive integerx≠0x\ne 0all x≠0x\ne 0quotient rule applied to 1/xm1/x^m
r=p/qr=p/q in lowest terms, non-integer, qq odd (so q≥3q\ge3)all x∈Rx\in\mathbb{R} if p>0p>0; only x≠0x\ne0 if p<0p<0x≠0x\ne 0 (and also at x=0x=0 when r>1r>1)implicit differentiation of yq=xpy^q=x^p
r=p/qr=p/q in lowest terms, qq even (never an integer)x≥0x\ge 0 if p>0p>0; only x>0x>0 if p<0p<0x>0x>0 (plus a one-sided derivative at x=0x=0 when r>1r>1)implicit differentiation of yq=xpy^q=x^p
rr irrationalx>0x>0x>0x>0xr=erln⁡xx^r=e^{r\ln x} (later in the course)

The third column lists where the formula is guaranteed; the parenthetical remarks record the extra points that are easy to lose. For example ddxx5/3=53x2/3\frac{d}{dx}x^{5/3}=\frac53x^{2/3} is valid at x=0x=0 as well (both sides are 00 there), whereas ddxx2/3=23x−1/3\frac{d}{dx}x^{2/3}=\frac23x^{-1/3} genuinely fails at x=0x=0, where the graph has a cusp.

Derivation sketch for a positive integer nn. Using bn−an=(b−a)(bn−1+bn−2a+⋯+an−1)b^n-a^n=(b-a)\left(b^{n-1}+b^{n-2}a+\cdots+a^{n-1}\right) with b=x+hb=x+h, a=xa=x, (x+h)n−xnh=(x+h)n−1+(x+h)n−2x+⋯+xn−1 →h→0 xn−1+⋯+xn−1⏟n terms=n xn−1.\frac{(x+h)^n-x^n}{h}=(x+h)^{n-1}+(x+h)^{n-2}x+\cdots+x^{n-1}\ \xrightarrow[h\to 0]{}\ \underbrace{x^{n-1}+\cdots+x^{n-1}}_{n\ \text{terms}}=n\,x^{n-1}.

Edge cases you must name.

  • x=x1/2\sqrt{x}=x^{1/2} has domain [0,∞)[0,\infty) but ddxx=12x\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}} only for x>0x>0. At x=0x=0 the one-sided difference quotient is hh=1h→+∞\frac{\sqrt{h}}{h}=\frac{1}{\sqrt h}\to+\infty, so x\sqrt{x} is not differentiable at 00.
  • x1/3x^{1/3} is defined for all real xx, but ddxx1/3=13x−2/3\frac{d}{dx}x^{1/3}=\frac13 x^{-2/3} requires x≠0x\ne 0; at x=0x=0 the graph has a vertical tangent and no derivative.
  • The power rule needs a variable base and a constant exponent. It does not apply to 2x2^{x}, and ddx2x≠x 2x−1\frac{d}{dx}2^{x}\ne x\,2^{x-1}.

Useful special cases: ddx(1x)=−1x2\dfrac{d}{dx}\left(\dfrac1x\right)=-\dfrac{1}{x^2}, ddx(1xn)=−nxn+1\dfrac{d}{dx}\left(\dfrac{1}{x^n}\right)=-\dfrac{n}{x^{n+1}}, ddxx=12x\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt x}, ddx(1x)=−12x3/2\dfrac{d}{dx}\left(\dfrac{1}{\sqrt x}\right)=-\dfrac{1}{2x^{3/2}}.

The product rule

Theorem. If ff and gg are differentiable at aa, then fgfg is differentiable at aa and (fg)′(a)=f′(a) g(a)+f(a) g′(a).(fg)'(a)=f'(a)\,g(a)+f(a)\,g'(a).

Derivation sketch. Add and subtract f(a)g(a+h)f(a)g(a+h) in the numerator:

\frac{f(a+h)g(a+h)-f(a)g(a)}{h}&=\frac{f(a+h)g(a+h)-f(a)g(a+h)+f(a)g(a+h)-f(a)g(a)}{h}\\ &=\frac{f(a+h)-f(a)}{h}\cdot g(a+h)+f(a)\cdot\frac{g(a+h)-g(a)}{h}. \end{aligned}$$ Let $h\to0$. The two difference quotients tend to $f'(a)$ and $g'(a)$ by hypothesis, and $g(a+h)\to g(a)$ because $g$, being differentiable at $a$, is continuous at $a$. That continuity step is exactly where the hypothesis on $g$ is used. For three factors, apply the rule twice: $(fgh)'=f'gh+fg'h+fgh'$ — differentiate one factor at a time and leave the others alone. ### The quotient rule **Theorem.** If $f$ and $g$ are differentiable at $a$ and $g(a)\ne 0$, then $f/g$ is differentiable at $a$ and $$\left(\frac{f}{g}\right)'(a)=\frac{f'(a)\,g(a)-f(a)\,g'(a)}{\big(g(a)\big)^{2}}.$$ The hypothesis $g(a)\ne0$ is essential: it also guarantees, by continuity of $g$, that $g\ne0$ on a whole interval around $a$, so the quotient is defined near $a$. *Derivation sketch.* $$\begin{aligned} \frac1h\left(\frac{f(a+h)}{g(a+h)}-\frac{f(a)}{g(a)}\right)&=\frac{f(a+h)g(a)-f(a)g(a+h)}{h\,g(a+h)g(a)}\\ &=\frac{g(a)\big[f(a+h)-f(a)\big]-f(a)\big[g(a+h)-g(a)\big]}{h\,g(a+h)g(a)}\\ &=\frac{\dfrac{f(a+h)-f(a)}{h}\,g(a)-f(a)\,\dfrac{g(a+h)-g(a)}{h}}{g(a+h)g(a)}, \end{aligned}$$ where the second line came from adding and subtracting $f(a)g(a)$ in the numerator. Letting $h\to0$ and using $g(a+h)\to g(a)\ne0$ gives the formula. Taking $f\equiv 1$ gives the **reciprocal rule**: $\left(\dfrac{1}{g}\right)'=-\dfrac{g'}{g^{2}}$ wherever $g\ne0$. Order matters in the numerator: it is $(\text{derivative of top})(\text{bottom})-(\text{top})(\text{derivative of bottom})$. Swapping the two terms flips every sign. ### Polynomials and rational functions A polynomial $P(x)=c_nx^n+\cdots+c_1x+c_0$ is differentiable on all of $\mathbb{R}$, with $$P'(x)=nc_nx^{n-1}+(n-1)c_{n-1}x^{n-2}+\cdots+c_1,$$ a polynomial of degree $n-1$ when $n\ge1$ and $c_n\ne0$. A rational function $R=P/Q$ is differentiable at every point of its domain (every $x$ with $Q(x)\ne0$), and $$R'=\frac{P'Q-PQ'}{Q^{2}}$$ is again a rational function whose denominator is $Q^2$. Useful check on degrees: if $\deg P=p\ge1$ and $\deg Q=q\ge1$, then $P'Q$ and $PQ'$ each have degree $p+q-1$, so $P'Q-PQ'$ has degree **at most** $p+q-1$. Writing $P=ax^{p}+\cdots$ and $Q=bx^{q}+\cdots$ with $ab\ne0$, the coefficient of $x^{p+q-1}$ in $P'Q-PQ'$ is $pab-qab=(p-q)ab$, so the leading terms cancel **exactly when $p=q$**, not in general. For instance $P=x^{2}$, $Q=x$ give $P'Q-PQ'=2x\cdot x-x^{2}\cdot 1=x^{2}$, of the full degree $p+q-1=2$; while $P=x^{2}+1$, $Q=x^{2}$ give $P'Q-PQ'=-2x$, of degree $1<3$. ### Worked example 1 — linearity plus the power rule with negative and fractional exponents Differentiate $f(x)=5x^{4}-\dfrac{3}{x^{2}}+7\sqrt{x}-9$. **Step 1 — rewrite every term as a power of $x$.** $$f(x)=5x^{4}-3x^{-2}+7x^{1/2}-9.$$ The natural domain is $x>0$ (the $\sqrt x$ needs $x\ge0$, the $x^{-2}$ needs $x\ne0$). **Step 2 — differentiate term by term (linearity), each term by the power rule.** $$\begin{aligned} \frac{d}{dx}\big(5x^4\big)&=5\cdot 4x^{3}=20x^{3},\\ \frac{d}{dx}\big(-3x^{-2}\big)&=-3\cdot(-2)x^{-3}=6x^{-3},\\ \frac{d}{dx}\big(7x^{1/2}\big)&=7\cdot\tfrac12 x^{-1/2}=\tfrac72 x^{-1/2},\\ \frac{d}{dx}(-9)&=0. \end{aligned}$$ **Step 3 — assemble and tidy.** $$f'(x)=20x^{3}+\frac{6}{x^{3}}+\frac{7}{2\sqrt{x}},\qquad x>0.$$ Note the sign: differentiating $-3x^{-2}$ produced a $+$, because $-3\times(-2)=+6$. ### Worked example 2 — the quotient rule, in full Differentiate $F(x)=\dfrac{2x^{3}-x}{x^{2}+1}$. **Step 1 — name the pieces.** $u=2x^{3}-x$, $v=x^{2}+1$, so $u'=6x^{2}-1$ and $v'=2x$. Since $x^2+1\ge1>0$ for all real $x$, the hypothesis $v\ne0$ holds everywhere. **Step 2 — apply the rule.** $$F'(x)=\frac{u'v-uv'}{v^{2}}=\frac{(6x^{2}-1)(x^{2}+1)-(2x^{3}-x)(2x)}{(x^{2}+1)^{2}}.$$ **Step 3 — expand the numerator carefully.** $$\begin{aligned} (6x^{2}-1)(x^{2}+1)&=6x^{4}+6x^{2}-x^{2}-1=6x^{4}+5x^{2}-1,\\ (2x^{3}-x)(2x)&=4x^{4}-2x^{2},\\ \text{numerator}&=\big(6x^{4}+5x^{2}-1\big)-\big(4x^{4}-2x^{2}\big)=2x^{4}+7x^{2}-1. \end{aligned}$$ The minus sign multiplies the *whole* second product — that is where most sign errors happen. **Step 4 — state the answer and sanity-check it.** $$F'(x)=\frac{2x^{4}+7x^{2}-1}{(x^{2}+1)^{2}}\quad\text{for all }x\in\mathbb{R}.$$ Check at $x=1$: $F'(1)=\frac{2+7-1}{4}=2$. A numerical slope check gives $\frac{F(1.001)-F(0.999)}{0.002}=1.9999992\ldots$, which rounds to $2.00000$ and so agrees with the computed value $F'(1)=2$. ### Simplify before differentiating The quotient rule is never wrong, but it is often the slowest route. Rewrite first whenever the denominator is a single power of $x$, or when the fraction cancels. | Expression | Better move | Result | |---|---|---| | $\dfrac{x^{4}-2x^{2}}{x}$ | divide term by term: $x^{3}-2x$ | $3x^{2}-2$ | | $\dfrac{x^{2}-3x}{\sqrt{x}}$ | rewrite as $x^{3/2}-3x^{1/2}$ | $\tfrac32x^{1/2}-\tfrac32x^{-1/2}$ | | $\dfrac{x^{2}-9}{x-3}$ | cancel: $x+3$ for $x\ne3$ | $1$ (for $x\ne 3$) | | $(x^{2}+1)(x^{3}-4x)$ | expand: $x^{5}-3x^{3}-4x$ | $5x^{4}-9x^{2}-4$ | | $\dfrac{3x-1}{x^{2}+2}$ | nothing simplifies — use the quotient rule | $\dfrac{-3x^{2}+2x+6}{(x^{2}+2)^{2}}$ | Two warnings. First, cancelling changes the domain: $\frac{x^2-9}{x-3}$ is undefined at $x=3$, so its derivative is $1$ on $x \ne 3$ and undefined at $x=3$. Second, "simplify first" is a tactic, not a rule — when the denominator has two or more terms, the quotient rule is usually the right tool. ### Common mistakes | Wrong | Right | Why | |---|---|---| | $(fg)'=f'g'$ | $(fg)'=f'g+fg'$ | Test $f=g=x$: $(x^2)'=2x$, not $1\cdot1=1$. | | $\left(\dfrac fg\right)'=\dfrac{f'}{g'}$ | $\dfrac{f'g-fg'}{g^{2}}$ | Test $f=x^2$, $g=x$: true answer $1$, the fake rule gives $2x$. | | $\dfrac{d}{dx}\dfrac{x^{2}+1}{x^{3}}=\dfrac{2x}{3x^{2}}$ | $-\dfrac{x^{2}+3}{x^{4}}$ | Same "top over bottom" error; here rewriting as $x^{-1}+x^{-3}$ is fastest. | | $\dfrac{f'g-fg'}{g^2}$ written as $\dfrac{fg'-f'g}{g^2}$ | $f'g$ comes first | Reversing the order negates the whole answer. | | $\dfrac{d}{dx}\big(3x^{-2}\big)=-6x^{-1}$ | $-6x^{-3}$ | The new exponent is $-2-1=-3$; subtract $1$, never add. | | $\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2}x^{1/2}$ | $\dfrac12 x^{-1/2}$ | $\tfrac12-1=-\tfrac12$. | | $\dfrac{d}{dx}2^{x}=x\,2^{x-1}$ | not the power rule | Constant base, variable exponent — the power rule does not apply. | | $\dfrac{d}{dx}\left(\dfrac{5}{x^{2}}\right)=\dfrac{5}{2x}$ | $-\dfrac{10}{x^{3}}$ | Rewrite as $5x^{-2}$ first; you cannot differentiate numerator and denominator separately. | | Forgetting $(x^2+1)^2$ stays squared | keep $g^2$ intact | Expanding the denominator hides the factor you need later for solving $f'=0$. | | $\dfrac{d}{dx}(x^{2}+1)(x-3)=2x(1)$ | $2x(x-3)+(x^2+1)$ | Differentiating both factors at once is the product-rule error again. | ### Fast self-checks - **Degree check.** Differentiating a polynomial of degree $n\ge1$ must lower the degree by exactly one. (For $n=0$ the derivative is the zero polynomial, whose degree is left undefined, so the check does not apply.) - **Numeric check.** Compute $\frac{f(a+0.001)-f(a-0.001)}{0.002}$ and compare with $f'(a)$; they should agree to about three decimals. - **Zero check.** For $f'=0$ from a quotient, only the numerator can vanish — solutions must still lie in the domain of $f$.

Key terms

  • Linearity of the derivative
  • Constant multiple rule
  • Sum and difference rule
  • Power rule
  • Negative integer exponent
  • Rational exponent
  • Product rule
  • Quotient rule
  • Reciprocal rule
  • Derivative of a polynomial
  • Rational function
  • Simplify before differentiating
  • Horizontal tangent
  • Domain restriction

Practice Problems

Differentiate f(x)=7x5−4x3+2x−9f(x)=7x^{5}-4x^{3}+2x-9, and evaluate f′(1)f'(1).

Show hint

Differentiation is linear, so handle one term at a time; the derivative of a constant is 00.

Show answer

Step 1 — use linearity. Because ff is a finite sum of constant multiples of powers, f′(x)=7ddx(x5)−4ddx(x3)+2ddx(x)−ddx(9).f'(x)=7\frac{d}{dx}\big(x^{5}\big)-4\frac{d}{dx}\big(x^{3}\big)+2\frac{d}{dx}\big(x\big)-\frac{d}{dx}\big(9\big).

Step 2 — apply the power rule to each term. ddxx5=5x4,ddxx3=3x2,ddxx=1,ddx(9)=0.\frac{d}{dx}x^{5}=5x^{4},\qquad \frac{d}{dx}x^{3}=3x^{2},\qquad \frac{d}{dx}x=1,\qquad \frac{d}{dx}(9)=0.

Step 3 — assemble. f′(x)=7(5x4)−4(3x2)+2(1)−0=35x4−12x2+2.f'(x)=7(5x^{4})-4(3x^{2})+2(1)-0=35x^{4}-12x^{2}+2.

Step 4 — evaluate. f′(1)=35−12+2=25.f'(1)=35-12+2=25.

Check. The original polynomial has degree 55 and the derivative has degree 44, as it must. Numerically, f(1.001)−f(0.999)0.002≈25.00\frac{f(1.001)-f(0.999)}{0.002}\approx 25.00, matching f′(1)=25f'(1)=25.

Answer: f′(x)=35x4−12x2+2f'(x)=35x^{4}-12x^{2}+2 and f′(1)=25f'(1)=25.

Differentiate g(x)=3x+5x−1x3g(x)=3\sqrt{x}+\dfrac{5}{x}-\dfrac{1}{x^{3}}. State the set of xx on which your formula is valid.

Show hint

Before differentiating anything, rewrite each term in the form c xrc\,x^{r}.

Show answer

Step 1 — rewrite as powers. g(x)=3x1/2+5x−1−x−3.g(x)=3x^{1/2}+5x^{-1}-x^{-3}.

Step 2 — differentiate term by term with the power rule.

\frac{d}{dx}\big(3x^{1/2}\big)&=3\cdot\tfrac12x^{-1/2}=\tfrac32x^{-1/2},\\ \frac{d}{dx}\big(5x^{-1}\big)&=5\cdot(-1)x^{-2}=-5x^{-2},\\ \frac{d}{dx}\big(-x^{-3}\big)&=-1\cdot(-3)x^{-4}=3x^{-4}. \end{aligned}$$ Notice the last sign: $(-1)\times(-3)=+3$. **Step 3 — assemble, in radical/fraction form.** $$g'(x)=\frac{3}{2\sqrt{x}}-\frac{5}{x^{2}}+\frac{3}{x^{4}}.$$ **Step 4 — domain.** The term $\sqrt x$ requires $x\ge0$ and the terms $5/x$, $1/x^{3}$ require $x\ne0$, so $g$ has domain $x>0$. The derivative formula is valid for all $x>0$ (at $x=0$ the function is not even defined, and $\sqrt x$ has no derivative there in any case). **Check.** At $x=1$: $g'(1)=\tfrac32-5+3=-\tfrac12$. Numerically $\frac{g(1.001)-g(0.999)}{0.002}\approx-0.5000$. **Answer:** $g'(x)=\dfrac{3}{2\sqrt{x}}-\dfrac{5}{x^{2}}+\dfrac{3}{x^{4}}$ for $x>0$.

Let h(x)=(x2+1)(x3−4x)h(x)=(x^{2}+1)(x^{3}-4x). Differentiate hh using the product rule, then verify your answer by multiplying out first and differentiating the resulting polynomial.

Show hint

Label f=x2+1f=x^2+1 and g=x3−4xg=x^3-4x and write down f′f' and g′g' before touching the rule.

Show answer

Method 1 — product rule.

Set f(x)=x2+1f(x)=x^{2}+1 and g(x)=x3−4xg(x)=x^{3}-4x. Then f′(x)=2xf'(x)=2x and g′(x)=3x2−4g'(x)=3x^{2}-4. Both are polynomials, hence differentiable everywhere, so the product rule applies for all xx: h′(x)=f′g+fg′=2x(x3−4x)+(x2+1)(3x2−4).h'(x)=f'g+fg'=2x\big(x^{3}-4x\big)+\big(x^{2}+1\big)\big(3x^{2}-4\big). Expand each piece:

2x\big(x^{3}-4x\big)&=2x^{4}-8x^{2},\\ \big(x^{2}+1\big)\big(3x^{2}-4\big)&=3x^{4}-4x^{2}+3x^{2}-4=3x^{4}-x^{2}-4. \end{aligned}$$ Adding, $$h'(x)=\big(2x^{4}-8x^{2}\big)+\big(3x^{4}-x^{2}-4\big)=5x^{4}-9x^{2}-4.$$ **Method 2 — expand first.** $$h(x)=(x^{2}+1)(x^{3}-4x)=x^{5}-4x^{3}+x^{3}-4x=x^{5}-3x^{3}-4x,$$ so by linearity and the power rule $$h'(x)=5x^{4}-9x^{2}-4.$$ The two methods agree. (A numerical check at $x=2$ gives $\frac{h(2.001)-h(1.999)}{0.002}\approx 40.00$, and $5(16)-9(4)-4=40$.) **Answer:** $h'(x)=5x^{4}-9x^{2}-4$.

Differentiate f(x)=3x−1x2+2f(x)=\dfrac{3x-1}{x^{2}+2} and find the slope of the tangent line at x=0x=0.

Show hint

The denominator is a sum of two terms, so nothing cancels — use the quotient rule and keep the numerator's subtraction order straight.

Show answer

Step 1 — name the pieces and check the hypothesis. Let u=3x−1u=3x-1 and v=x2+2v=x^{2}+2, so u′=3u'=3 and v′=2xv'=2x. Since x2+2≥2>0x^{2}+2\ge2>0 for every real xx, we have v≠0v\ne0 everywhere and the quotient rule applies on all of R\mathbb{R}.

Step 2 — apply the quotient rule. f′(x)=u′v−uv′v2=3(x2+2)−(3x−1)(2x)(x2+2)2.f'(x)=\frac{u'v-uv'}{v^{2}}=\frac{3\big(x^{2}+2\big)-(3x-1)(2x)}{\big(x^{2}+2\big)^{2}}.

Step 3 — expand the numerator.

3\big(x^{2}+2\big)&=3x^{2}+6,\\ (3x-1)(2x)&=6x^{2}-2x,\\ \text{numerator}&=\big(3x^{2}+6\big)-\big(6x^{2}-2x\big)=-3x^{2}+2x+6. \end{aligned}$$ The minus sign distributes over both terms of $6x^{2}-2x$; that is the step to be careful about. **Step 4 — the derivative.** $$f'(x)=\frac{-3x^{2}+2x+6}{\big(x^{2}+2\big)^{2}},\qquad x\in\mathbb{R}.$$ **Step 5 — slope at $x=0$.** $$f'(0)=\frac{-0+0+6}{(0+2)^{2}}=\frac{6}{4}=\frac32.$$ **Check.** To eight decimals, $f(0.001)\approx-0.49849975$ and $f(-0.001)\approx-0.50149975$, so $$\frac{f(0.001)-f(-0.001)}{0.002}\approx\frac{0.00300000}{0.002}=1.50000,$$ matching $\tfrac32$. (Carried out in full precision the quotient is $1.4999992\ldots$. The two function values are both close to $-0.5$ and differ by only $0.003$, so they must be kept to many digits before subtracting; rounding them to three or four decimals first would destroy the check.) **Answer:** $f'(x)=\dfrac{-3x^{2}+2x+6}{\big(x^{2}+2\big)^{2}}$, and the tangent slope at $x=0$ is $\dfrac32$.

Differentiate f(x)=x4−2x2+xx2f(x)=\dfrac{x^{4}-2x^{2}+\sqrt{x}}{x^{2}} without using the quotient rule. State the domain.

Show hint

A single power of xx in the denominator means you can divide term by term and use the law xa/xb=xa−bx^{a}/x^{b}=x^{a-b}.

Show answer

Step 1 — split the fraction. Dividing each term of the numerator by x2x^{2}: f(x)=x4x2−2x2x2+x1/2x2=x2−2+x1/2−2=x2−2+x−3/2.f(x)=\frac{x^{4}}{x^{2}}-\frac{2x^{2}}{x^{2}}+\frac{x^{1/2}}{x^{2}}=x^{2}-2+x^{1/2-2}=x^{2}-2+x^{-3/2}.

Step 2 — domain. The x\sqrt x forces x≥0x\ge0 and the division by x2x^{2} forces x≠0x\ne0, so the domain is x>0x>0.

Step 3 — differentiate term by term.

\frac{d}{dx}\big(x^{2}\big)&=2x,\\ \frac{d}{dx}(-2)&=0,\\ \frac{d}{dx}\big(x^{-3/2}\big)&=-\tfrac32x^{-5/2}. \end{aligned}$$ **Step 4 — assemble.** $$f'(x)=2x-\frac{3}{2}x^{-5/2}=2x-\frac{3}{2x^{5/2}},\qquad x>0.$$ **Check.** At $x=1$: $f'(1)=2-\tfrac32=\tfrac12$. Numerically $\frac{f(1.001)-f(0.999)}{0.002}\approx0.5000$. At $x=4$: $f'(4)=8-\frac{3}{2\cdot32}=8-\frac{3}{64}=7.953125$, and the numerical slope agrees to four decimals. **Answer:** $f'(x)=2x-\dfrac{3}{2x^{5/2}}$ on $x>0$.

Differentiate y=x2(2x−1)(x+3)y=x^{2}(2x-1)(x+3) by applying the product rule to the three factors, and confirm the result by expanding first.

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For three factors, differentiate one factor at a time and leave the other two alone; there are three such terms.

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Step 1 — the three-factor product rule. Applying the two-factor rule twice to uvwu v w gives (uvw)′=u′vw+uv′w+uvw′.(uvw)'=u'vw+uv'w+uvw'. With u=x2u=x^{2}, v=2x−1v=2x-1, w=x+3w=x+3 we have u′=2xu'=2x, v′=2v'=2, w′=1w'=1, so y′=2x(2x−1)(x+3)+x2⋅2⋅(x+3)+x2(2x−1)⋅1.y'=2x(2x-1)(x+3)+x^{2}\cdot 2\cdot(x+3)+x^{2}(2x-1)\cdot 1.

Step 2 — expand each term.

(2x-1)(x+3)&=2x^{2}+6x-x-3=2x^{2}+5x-3,\\ 2x\big(2x^{2}+5x-3\big)&=4x^{3}+10x^{2}-6x,\\ 2x^{2}(x+3)&=2x^{3}+6x^{2},\\ x^{2}(2x-1)&=2x^{3}-x^{2}. \end{aligned}$$ **Step 3 — add the three terms.** $$\begin{aligned} y'&=\big(4x^{3}+10x^{2}-6x\big)+\big(2x^{3}+6x^{2}\big)+\big(2x^{3}-x^{2}\big)\\ &=(4+2+2)x^{3}+(10+6-1)x^{2}-6x\\ &=8x^{3}+15x^{2}-6x. \end{aligned}$$ **Step 4 — confirm by expanding first.** $$y=x^{2}\big(2x^{2}+5x-3\big)=2x^{4}+5x^{3}-3x^{2}\ \Longrightarrow\ y'=8x^{3}+15x^{2}-6x.$$ The two agree. (Numerically at $x=2$: $\frac{y(2.001)-y(1.999)}{0.002}\approx112.00$, and $8(8)+15(4)-12=112$.) **Answer:** $y'=8x^{3}+15x^{2}-6x=x\big(8x^{2}+15x-6\big)$.

A student writes ddx[x2+1x3]=2x3x2.\frac{d}{dx}\left[\frac{x^{2}+1}{x^{3}}\right]=\frac{2x}{3x^{2}}. Explain precisely what is wrong, then compute the correct derivative in two independent ways and check that they agree.

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Ask which rule, if any, allows you to differentiate a numerator and a denominator separately.

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The error. The student differentiated the top and the bottom separately, i.e. used the non-existent rule (fg)′=f′g′\left(\frac{f}{g}\right)'=\frac{f'}{g'}. That rule is false. A one-line refutation: take f=x2f=x^{2}, g=xg=x. Then f/g=xf/g=x, whose derivative is 11, while f′/g′=2x1=2xf'/g'=\frac{2x}{1}=2x. (The fake rule is a garbled memory of l'Hopital's rule, and even that rule is not "differentiate top and bottom" without conditions. L'Hopital's rule says: if ff and gg are differentiable on a punctured interval around aa with g′(x)≠0g'(x)\ne0 there, if f(x)g(x)\frac{f(x)}{g(x)} is an indeterminate form 00\frac{0}{0} or ∞∞\frac{\infty}{\infty} as x→ax\to a, and if lim⁡x→af′(x)g′(x)\lim_{x\to a}\frac{f'(x)}{g'(x)} exists or is ±∞\pm\infty, then lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}. Its conclusion is about a limit of a quotient, never about the derivative of a quotient.)

Correct method 1 — rewrite, then use the power rule. x2+1x3=x2x3+1x3=x−1+x−3,\frac{x^{2}+1}{x^{3}}=\frac{x^{2}}{x^{3}}+\frac{1}{x^{3}}=x^{-1}+x^{-3}, so for x≠0x\ne0 ddx(x−1+x−3)=−x−2−3x−4=−1x2−3x4=−x2+3x4.\frac{d}{dx}\big(x^{-1}+x^{-3}\big)=-x^{-2}-3x^{-4}=-\frac{1}{x^{2}}-\frac{3}{x^{4}}=-\frac{x^{2}+3}{x^{4}}.

Correct method 2 — the quotient rule. With u=x2+1u=x^{2}+1, v=x3v=x^{3}, u′=2xu'=2x, v′=3x2v'=3x^{2}, valid wherever v≠0v\ne0, i.e. x≠0x\ne0:

\frac{d}{dx}\left[\frac{x^{2}+1}{x^{3}}\right]&=\frac{2x\cdot x^{3}-\big(x^{2}+1\big)\cdot 3x^{2}}{\big(x^{3}\big)^{2}}\\ &=\frac{2x^{4}-3x^{4}-3x^{2}}{x^{6}}\\ &=\frac{-x^{4}-3x^{2}}{x^{6}}\\ &=\frac{-x^{2}\big(x^{2}+3\big)}{x^{6}}=-\frac{x^{2}+3}{x^{4}}. \end{aligned}$$ The two methods give the same function, so the answer is confirmed. Method 1 is clearly faster whenever the denominator is a single power of $x$. **How wrong was the student?** At $x=1$ the correct slope is $-\frac{1+3}{1}=-4$, while the student's expression gives $\frac{2}{3}$ — not even the right sign. A numerical check confirms $\frac{f(1.001)-f(0.999)}{0.002}\approx-4.000$. **Answer:** $\dfrac{d}{dx}\left[\dfrac{x^{2}+1}{x^{3}}\right]=-\dfrac{x^{2}+3}{x^{4}}$ for $x\ne 0$.

The functions ff and gg are differentiable, with the values below.

fff′f'ggg′g'
at x=2x=233−1-1−4-455

Compute each of the following at x=2x=2: (a) (3f−2g)′(2)(3f-2g)'(2); (b) (fg)′(2)(fg)'(2); (c) (fg)′(2)\left(\dfrac{f}{g}\right)'(2); (d) (gf)′(2)\left(\dfrac{g}{f}\right)'(2); (e) ddx[x2f(x)]\dfrac{d}{dx}\big[x^{2}f(x)\big] at x=2x=2.

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You never need formulas for ff and gg — only the rule that matches the combination, evaluated at the single point x=2x=2.

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Throughout, f(2)=3f(2)=3, f′(2)=−1f'(2)=-1, g(2)=−4g(2)=-4, g′(2)=5g'(2)=5.

(a) Linearity. (3f−2g)′(2)=3f′(2)−2g′(2)=3(−1)−2(5)=−3−10=−13.(3f-2g)'(2)=3f'(2)-2g'(2)=3(-1)-2(5)=-3-10=-13.

(b) Product rule. (fg)′(2)=f′(2)g(2)+f(2)g′(2)=(−1)(−4)+(3)(5)=4+15=19.(fg)'(2)=f'(2)g(2)+f(2)g'(2)=(-1)(-4)+(3)(5)=4+15=19.

(c) Quotient rule. The hypothesis g(2)=−4≠0g(2)=-4\ne0 holds, so (fg)′(2)=f′(2)g(2)−f(2)g′(2)(g(2))2=(−1)(−4)−(3)(5)(−4)2=4−1516=−1116.\left(\frac{f}{g}\right)'(2)=\frac{f'(2)g(2)-f(2)g'(2)}{\big(g(2)\big)^{2}}=\frac{(-1)(-4)-(3)(5)}{(-4)^{2}}=\frac{4-15}{16}=-\frac{11}{16}.

(d) Quotient rule with the roles swapped. Here the hypothesis is f(2)=3≠0f(2)=3\ne0: (gf)′(2)=g′(2)f(2)−g(2)f′(2)(f(2))2=(5)(3)−(−4)(−1)32=15−49=119.\left(\frac{g}{f}\right)'(2)=\frac{g'(2)f(2)-g(2)f'(2)}{\big(f(2)\big)^{2}}=\frac{(5)(3)-(-4)(-1)}{3^{2}}=\frac{15-4}{9}=\frac{11}{9}. Note that (d) is not the reciprocal of (c): −1116-\frac{11}{16} and 119\frac{11}{9} are unrelated in that way. (In fact (gf)′=−(f/g)′(f/g)2\left(\frac gf\right)'=-\frac{(f/g)'}{(f/g)^2}, and indeed −(−1116)/(3−4)2=1116⋅169=119-\left(-\frac{11}{16}\right)\big/\left(\frac{3}{-4}\right)^{2}=\frac{11}{16}\cdot\frac{16}{9}=\frac{11}{9}, which confirms both answers at once.)

(e) Product rule with u=x2u=x^{2}. ddx[x2f(x)]=2x f(x)+x2f′(x) ⟹ at x=2: 2(2)(3)+(4)(−1)=12−4=8.\frac{d}{dx}\big[x^{2}f(x)\big]=2x\,f(x)+x^{2}f'(x)\ \Longrightarrow\ \text{at }x=2:\ 2(2)(3)+(4)(-1)=12-4=8.

Answers: (a) −13-13; (b) 1919; (c) −1116-\dfrac{11}{16}; (d) 119\dfrac{11}{9}; (e) 88.

Find every point on the graph of f(x)=x2+3x−1f(x)=\dfrac{x^{2}+3}{x-1} at which the tangent line is horizontal.

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Horizontal tangent means f′(x)=0f'(x)=0; for a quotient, only the numerator of f′f' can vanish — and any solution must lie in the domain of ff.

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Step 1 — domain. ff is defined and differentiable for all x≠1x\ne1.

Step 2 — differentiate with the quotient rule. Let u=x2+3u=x^{2}+3, v=x−1v=x-1, so u′=2xu'=2x, v′=1v'=1: f′(x)=2x(x−1)−(x2+3)(1)(x−1)2.f'(x)=\frac{2x(x-1)-\big(x^{2}+3\big)(1)}{(x-1)^{2}}.

Step 3 — simplify the numerator. 2x(x−1)−(x2+3)=2x2−2x−x2−3=x2−2x−3=(x−3)(x+1).2x(x-1)-\big(x^{2}+3\big)=2x^{2}-2x-x^{2}-3=x^{2}-2x-3=(x-3)(x+1). Hence f′(x)=(x−3)(x+1)(x−1)2,x≠1.f'(x)=\frac{(x-3)(x+1)}{(x-1)^{2}},\qquad x\ne1.

Step 4 — solve f′(x)=0f'(x)=0. A fraction is zero exactly when its numerator is zero and its denominator is not: (x−3)(x+1)=0  ⟺  x=3 or x=−1,(x-3)(x+1)=0\iff x=3\ \text{or}\ x=-1, and both satisfy x≠1x\ne1, so both are legitimate.

Step 5 — get the points. f(3)=9+33−1=122=6,f(−1)=1+3−1−1=4−2=−2.f(3)=\frac{9+3}{3-1}=\frac{12}{2}=6,\qquad f(-1)=\frac{1+3}{-1-1}=\frac{4}{-2}=-2.

Check. f′(0)=(−3)(1)1=−3f'(0)=\frac{(-3)(1)}{1}=-3 and the numerical slope at 00 is −3.000-3.000; f′(5)=(2)(6)16=0.75f'(5)=\frac{(2)(6)}{16}=0.75 and the numerical slope at 55 is 0.7500.750. The numerical slopes at x=3x=3 and x=−1x=-1 are 00 to four decimals.

Answer: the tangent is horizontal at (3,6)(3,6) and at (−1,−2)(-1,-2), with tangent lines y=6y=6 and y=−2y=-2.

(a) Using only the quotient rule and the power rule for positive integers, prove that for every positive integer mm, ddx x−m=−m x−m−1,x≠0.\frac{d}{dx}\,x^{-m}=-m\,x^{-m-1},\qquad x\ne 0.

(b) Explain why this argument says nothing about ddx2x\dfrac{d}{dx}2^{x}, and give a concrete reason why ddx2x=x 2x−1\dfrac{d}{dx}2^{x}=x\,2^{x-1} must be false.

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For (a), write x−mx^{-m} as a quotient with numerator the constant function 11. For (b), look at which of the base and the exponent is the variable.

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(a) Fix a positive integer mm and write x−m=1xm,x^{-m}=\frac{1}{x^{m}}, which is legitimate for every x≠0x\ne0. Put u(x)=1u(x)=1 and v(x)=xmv(x)=x^{m}. Then u′(x)=0u'(x)=0 (derivative of a constant) and v′(x)=mxm−1v'(x)=mx^{m-1} by the power rule for positive integers, and v(x)=xm≠0v(x)=x^{m}\ne0 precisely because x≠0x\ne0 — so the hypotheses of the quotient rule are satisfied at every such xx. Applying it,

\frac{d}{dx}\,x^{-m}&=\frac{u'v-uv'}{v^{2}} =\frac{0\cdot x^{m}-1\cdot m x^{m-1}}{\big(x^{m}\big)^{2}}\\ &=\frac{-m x^{m-1}}{x^{2m}} =-m\,x^{(m-1)-2m}\\ &=-m\,x^{-m-1}. \end{aligned}$$ This is exactly the power rule with exponent $r=-m$, since $r-1=-m-1$. It holds for all $x\ne0$, positive or negative. $\blacksquare$ *Spot check:* $m=2$ gives $\frac{d}{dx}x^{-2}=-2x^{-3}$, and at $x=2$ that is $-\frac{2}{8}=-0.25$; numerically $\frac{(2.001)^{-2}-(1.999)^{-2}}{0.002}\approx-0.2500$. **(b)** In the proof, the variable $x$ appears as the **base** and the exponent $m$ is a fixed constant — that is what allowed the power rule for $x^{m}$ to be used. In $2^{x}$ the situation is reversed: the base $2$ is constant and the **exponent** is the variable, so $2^{x}$ is not of the form $x^{r}$ and neither the power rule nor the argument above applies to it. A concrete refutation of the proposed formula: $2^{x}$ is increasing (if $x_1<x_2$ then $2^{x_1}<2^{x_2}$), and an increasing function that is differentiable at a point has derivative $\ge0$ there — every difference quotient $\frac{2^{x+h}-2^{x}}{h}$ is positive, and a limit of positive quantities cannot be negative. The proposed formula $x\,2^{x-1}$ takes the value $-1\cdot 2^{-2}=-\tfrac14$ at $x=-1$, which is negative, so the formula is false. The other value is worth a second look, because it shows how much care a "proof by monotonicity" needs. The formula also predicts $0$ at $x=0$, i.e. a horizontal tangent — and *that* alone is not yet a contradiction: a strictly increasing differentiable function may perfectly well have derivative $0$ at an isolated point, as $x^{3}$ does at $x=0$. To rule this value out you need an estimate rather than monotonicity. The difference quotients do the job: $\frac{2^{0.001}-2^{-0.001}}{0.002}\approx0.6931$, nowhere near $0$. (The true derivative, established later in the course, is $\frac{d}{dx}2^{x}=2^{x}\ln 2$, which is positive everywhere; at $x=0$ it equals $\ln2\approx0.693$, matching that numerical slope.)

Let f(x)=xx2+1f(x)=\dfrac{x}{x^{2}+1}. Find f′f', determine all intervals on which ff is increasing and on which it is decreasing, and find the coordinates of the points with horizontal tangents.

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After the quotient rule, factor the numerator; the squared denominator can never change the sign.

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Step 1 — domain. x2+1≥1>0x^{2}+1\ge1>0 for all real xx, so ff is defined and differentiable on all of R\mathbb{R} and the quotient rule applies everywhere.

Step 2 — differentiate. With u=xu=x, v=x2+1v=x^{2}+1, u′=1u'=1, v′=2xv'=2x: f′(x)=1⋅(x2+1)−x⋅2x(x2+1)2=x2+1−2x2(x2+1)2=1−x2(x2+1)2.f'(x)=\frac{1\cdot\big(x^{2}+1\big)-x\cdot 2x}{\big(x^{2}+1\big)^{2}}=\frac{x^{2}+1-2x^{2}}{\big(x^{2}+1\big)^{2}}=\frac{1-x^{2}}{\big(x^{2}+1\big)^{2}}.

Step 3 — sign analysis. The denominator (x2+1)2\big(x^{2}+1\big)^{2} is strictly positive for every xx, so f′(x)f'(x) has the same sign as the numerator 1−x2=(1−x)(1+x).1-x^{2}=(1-x)(1+x). This is positive when −1<x<1-1<x<1, zero at x=±1x=\pm1, and negative when ∣x∣>1|x|>1:

intervalsign of 1−x21-x^2sign of f′f'behaviour of ff
(−∞,−1)(-\infty,-1)negativenegativedecreasing
(−1,1)(-1,1)positivepositiveincreasing
(1,∞)(1,\infty)negativenegativedecreasing

Step 4 — horizontal tangents. f′(x)=0f'(x)=0 exactly when 1−x2=01-x^{2}=0, i.e. x=1x=1 or x=−1x=-1. The corresponding values are f(1)=11+1=12,f(−1)=−11+1=−12.f(1)=\frac{1}{1+1}=\frac12,\qquad f(-1)=\frac{-1}{1+1}=-\frac12. So the horizontal tangents occur at (1,12)\left(1,\tfrac12\right) and (−1,−12)\left(-1,-\tfrac12\right), with tangent lines y=12y=\tfrac12 and y=−12y=-\tfrac12.

Check. f′(0)=1−01=1f'(0)=\frac{1-0}{1}=1 and the numerical slope at 00 is 1.0001.000; f′(2)=1−425=−0.12f'(2)=\frac{1-4}{25}=-0.12 and the numerical slope at 22 is −0.120-0.120. Signs match the table.

Answer: f′(x)=1−x2(x2+1)2f'(x)=\dfrac{1-x^{2}}{\big(x^{2}+1\big)^{2}}; ff is increasing on (−1,1)(-1,1) and decreasing on (−∞,−1)(-\infty,-1) and on (1,∞)(1,\infty); horizontal tangents at (−1,−12)\left(-1,-\tfrac12\right) and (1,12)\left(1,\tfrac12\right).

Let F(x)=x (x−1)x+1F(x)=\dfrac{\sqrt{x}\,(x-1)}{x+1}. Differentiate FF, express F′F' as a single fraction with no negative or fractional exponents in the numerator, state the domain of F′F', and find the xx-coordinate of the point where the tangent is horizontal.

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First rewrite the numerator x (x−1)\sqrt x\,(x-1) as a sum of powers of xx; then one quotient rule finishes it, and clearing x−1/2x^{-1/2} at the end is just multiplying top and bottom by 2x2\sqrt x.

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Step 1 — domain and setup. x\sqrt x requires x≥0x\ge0, and x+1>0x+1>0 there, so FF has domain [0,∞)[0,\infty); since x\sqrt x is not differentiable at 00, we work on x>0x>0.

Rewrite the numerator as a sum of powers: u(x)=x (x−1)=x1/2⋅x−x1/2=x3/2−x1/2,u(x)=\sqrt{x}\,(x-1)=x^{1/2}\cdot x-x^{1/2}=x^{3/2}-x^{1/2}, so by the power rule u′(x)=32x1/2−12x−1/2.u'(x)=\tfrac32x^{1/2}-\tfrac12x^{-1/2}. Let v(x)=x+1v(x)=x+1, so v′(x)=1v'(x)=1 and v(x)≠0v(x)\ne0 on x>0x>0.

Step 2 — quotient rule. F′(x)=(32x1/2−12x−1/2)(x+1)−(x3/2−x1/2)(x+1)2.F'(x)=\frac{\left(\tfrac32x^{1/2}-\tfrac12x^{-1/2}\right)(x+1)-\left(x^{3/2}-x^{1/2}\right)}{(x+1)^{2}}.

Step 3 — expand the numerator.

\left(\tfrac32x^{1/2}-\tfrac12x^{-1/2}\right)(x+1)&=\tfrac32x^{3/2}+\tfrac32x^{1/2}-\tfrac12x^{1/2}-\tfrac12x^{-1/2}\\ &=\tfrac32x^{3/2}+x^{1/2}-\tfrac12x^{-1/2}, \end{aligned}$$ using $x^{1/2}\cdot x=x^{3/2}$ and $x^{-1/2}\cdot x=x^{1/2}$. Subtracting $\left(x^{3/2}-x^{1/2}\right)$: $$\text{numerator}=\left(\tfrac32-1\right)x^{3/2}+(1+1)x^{1/2}-\tfrac12x^{-1/2}=\tfrac12x^{3/2}+2x^{1/2}-\tfrac12x^{-1/2}.$$ **Step 4 — clear the fractional powers.** Multiply numerator and denominator by $2x^{1/2}$ (allowed since $x>0$): $$2x^{1/2}\left(\tfrac12x^{3/2}+2x^{1/2}-\tfrac12x^{-1/2}\right)=x^{2}+4x-1,$$ so $$F'(x)=\frac{x^{2}+4x-1}{2\sqrt{x}\,(x+1)^{2}},\qquad x>0.$$ **Step 5 — horizontal tangent.** The denominator is positive on $x>0$, so $F'(x)=0$ exactly when $$x^{2}+4x-1=0\iff x=\frac{-4\pm\sqrt{16+4}}{2}=-2\pm\sqrt5 .$$ Only $x=-2+\sqrt5\approx0.2361$ lies in the domain $x>0$; the root $-2-\sqrt5$ is rejected. Note also that $F'>0$ for $x>-2+\sqrt5$ and $F'<0$ on $\left(0,-2+\sqrt5\right)$, so this is a minimum point. **Check.** At $x=1$: $F'(1)=\frac{1+4-1}{2\cdot1\cdot4}=\frac48=\frac12$, and the numerical slope $\frac{F(1.001)-F(0.999)}{0.002}\approx0.5000$. At $x=0.25$: $F'(0.25)=\frac{0.0625+1-1}{2(0.5)(1.5625)}=\frac{0.0625}{1.5625}=0.04$, and the numerical slope there is $0.0400$. **Answer:** $F'(x)=\dfrac{x^{2}+4x-1}{2\sqrt{x}\,(x+1)^{2}}$ for $x>0$, with a horizontal tangent at $x=-2+\sqrt5$.