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The Derivative & Its Rules · Topic 07 of 23

Basic Differentiation Rules

Differentiation is linear — (cf)′=cf′(cf)'=cf' and (f±g)′=f′±g′(f\pm g)'=f'\pm g' — and the power rule ddxxr=rxr−1\frac{d}{dx}x^r=rx^{r-1} handles every term cxrcx^r once roots and reciprocals are rewritten as powers. Products and quotients need their own rules, (fg)′=f′g+fg′(fg)'=f'g+fg' and (fg)′=f′g−fg′g2\left(\frac fg\right)'=\frac{f'g-fg'}{g^2}, because the derivative of a product is not the product of the derivatives.

4 min readCore — on every final6 formulas6 quick checks
01

Key ideas

5 things to remember
  1. 1

    Rewrite every term as c xrc\,x^r first

    x=x1/2\sqrt{x}=x^{1/2}, 5x2=5x−2\frac{5}{x^2}=5x^{-2}, 1x=x−1/2\frac{1}{\sqrt x}=x^{-1/2}. Once each term is a constant times a power, linearity and the power rule finish it: differentiate term by term, multiply by the exponent, subtract one from it.

  2. 2

    Power rule: variable base, constant exponent

    ddxxr=rxr−1\frac{d}{dx}x^r=rx^{r-1} for any real rr, wherever both sides make sense. It does not apply to 2x2^x (constant base, variable exponent), and ddxx=12x\frac{d}{dx}\sqrt x=\frac{1}{2\sqrt x} holds only for x>0x>0.

  3. 3

    Product rule: one factor at a time

    (fg)′=f′g+fg′(fg)'=f'g+fg': each term differentiates exactly one factor and leaves the other alone, so three factors give three terms, (uvw)′=u′vw+uv′w+uvw′(uvw)'=u'vw+uv'w+uvw'. Never (fg)′=f′g′(fg)'=f'g' — test it with f=g=xf=g=x.

  4. 4

    Quotient rule: order and the square

    (fg)′=f′g−fg′g2\left(\frac fg\right)'=\frac{f'g-fg'}{g^2} wherever g≠0g\ne0, with the derivative of the top first — swapping the two terms flips the sign of the whole answer. Keep the denominator as g2g^2, factored, never expanded.

  5. 5

    Simplify before you differentiate

    A single power of xx underneath means divide term by term (x4−2x2x=x3−2x\frac{x^4-2x^2}{x}=x^3-2x), and two short polynomial factors are faster multiplied out than product-ruled. Save the quotient rule for denominators with two or more terms.

-2.0-1.5-1.0-0.50.51.01.52.0-224xylocal maxlocal minf′(−1) = 0f′(1) = 0f(x) = x³ − 3xf′(x) = 3x² − 3
f(x)=x3−3xf(x)=x^3-3x (blue) and, from the power rule, f′(x)=3x2−3f'(x)=3x^2-3 (dashed amber). Where ff peaks or bottoms out, f′f' crosses zero: f′(±1)=0f'(\pm1)=0. Solving f′(x)=0f'(x)=0 is how horizontal tangents are found.
fΔfgΔgf · gg · Δff · ΔgΔf · Δg
Why the product rule has two terms. Stretch an f×gf\times g rectangle by Δf\Delta f and Δg\Delta g: the added area is the two shaded strips g Δf+f Δgg\,\Delta f+f\,\Delta g plus the small corner Δf Δg\Delta f\,\Delta g. Divide by Δx\Delta x and let Δx→0\Delta x\to0 — the strips become f′g+fg′f'g+fg' and the corner vanishes. Nothing here looks like f′g′f'g'.
02

Formulas

What to have memorised

  • Linearity

    (cf)′=c f′,(f±g)′=f′±g′(cf)'=c\,f',\qquad (f\pm g)'=f'\pm g'

    Constants ride along; sums go term by term.

  • Power rule

    ddx xr=r xr−1\frac{d}{dx}\,x^{r}=r\,x^{r-1}

    Any real rr, variable base. And ddx(c)=0\frac{d}{dx}(c)=0.

  • Product rule

    (fg)′=f′g+fg′(fg)'=f'g+fg'

    One factor differentiated per term. Three factors: u′vw+uv′w+uvw′u'vw+uv'w+uvw'.

  • Quotient rule

    (fg)′=f′g−fg′g2\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}}

    Needs g≠0g\neq0. Low d-high minus high d-low, over low squared.

  • Reciprocal rule

    (1g)′=−g′g2\left(\frac{1}{g}\right)'=-\frac{g'}{g^{2}}

    Quotient rule with f≡1f\equiv1; it gives ddxx−n=−nx−n−1\frac{d}{dx}x^{-n}=-nx^{-n-1}.

  • Two powers worth memorizing

    ddxx=12x,ddx ⁣(1x)=−1x2\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}},\qquad \frac{d}{dx}\!\left(\frac{1}{x}\right)=-\frac{1}{x^{2}}

    Power rule with r=12r=\frac12 (for x>0x>0) and r=−1r=-1 (for x≠0x\neq0).

03

Differentiate with the rules

The steps, in order
  1. 1

    Rewrite every term as c xrc\,x^{r}: x=x1/2\sqrt{x}=x^{1/2}, 5x2=5x−2\frac{5}{x^2}=5x^{-2}. Note the domain now, before anything cancels.

  2. 2

    Sums: go term by term. Power rule: multiply by the exponent, then subtract one from it. Constants become 00.

  3. 3

    Products: write ff, gg, f′f', g′g' first, then f′g+fg′f'g+fg'. Two short polynomial factors are often faster expanded.

  4. 4

    Quotients: a one-term denominator means divide through instead. Otherwise f′g−fg′g2\frac{f'g-fg'}{g^2} — top's derivative first, denominator kept squared.

  5. 5

    Simplify and factor the numerator. For horizontal tangents set it to 00 and discard any root outside the domain.

  6. 6

    Check: a degree-nn polynomial's derivative has degree n−1n-1, and f(a+0.001)−f(a−0.001)0.002\frac{f(a+0.001)-f(a-0.001)}{0.002} should match f′(a)f'(a).

04

Watch out

The mistakes that cost marks

  • ✗ Wrong

    (fg)′=f′g′(fg)'=f'g'

    ✓ Right

    (fg)′=f′g+fg′(fg)'=f'g+fg' — one factor differentiated per term.

    Why: Test f=g=xf=g=x: (x2)′=2x(x^2)'=2x, not 1⋅1=11\cdot1=1.

  • ✗ Wrong

    ddx ⁣[x2+1x3]=2x3x2\dfrac{d}{dx}\!\left[\dfrac{x^2+1}{x^3}\right]=\dfrac{2x}{3x^2}

    ✓ Right

    Rewrite as x−1+x−3x^{-1}+x^{-3}: the derivative is −x−2−3x−4=−x2+3x4-x^{-2}-3x^{-4}=-\dfrac{x^2+3}{x^4}.

    Why: There is no top-over-bottom rule; at x=1x=1 the fake answer even has the wrong sign.

  • ✗ Wrong

    (fg)′=fg′−f′gg2\left(\dfrac fg\right)'=\dfrac{fg'-f'g}{g^2}

    ✓ Right

    f′g−fg′g2\dfrac{f'g-fg'}{g^2}: the derivative of the top comes first.

    Why: Swapping the two terms negates the whole answer.

  • ✗ Wrong

    ddx(3x−2)=−6x−1\dfrac{d}{dx}\big(3x^{-2}\big)=-6x^{-1}

    ✓ Right

    −6x−3-6x^{-3}: the exponent drops from −2-2 to −3-3.

    Why: Subtract one from the exponent, never add.

  • ✗ Wrong

    ddxx=12x1/2\dfrac{d}{dx}\sqrt{x}=\dfrac12x^{1/2}

    ✓ Right

    12x−1/2=12x\dfrac12x^{-1/2}=\dfrac{1}{2\sqrt x}, valid for x>0x>0 only.

    Why: 12−1=−12\tfrac12-1=-\tfrac12; at x=0x=0 the tangent is vertical.

  • ✗ Wrong

    ddx2x=x 2x−1\dfrac{d}{dx}2^{x}=x\,2^{x-1}

    ✓ Right

    Not a power-rule case: constant base, variable exponent. (Later: ddx2x=2xln⁡2\frac{d}{dx}2^x=2^x\ln2.)

    Why: 2x2^x is increasing, yet x 2x−1x\,2^{x-1} is negative at x=−1x=-1.

05

Quick check

Commit to an answer before you reveal one
  1. Q1easy

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

    Hint

    Rewrite every term as c xrc\,x^r before differentiating anything.

    Show answer

    Answer

    g′(x)=32x−5x2+3x4g'(x)=\dfrac{3}{2\sqrt{x}}-\dfrac{5}{x^2}+\dfrac{3}{x^4} for x>0x>0.

    Steps

    Rewrite: g(x)=3x1/2+5x−1−x−3g(x)=3x^{1/2}+5x^{-1}-x^{-3}.

    g′(x)=32x−1/2−5x−2+3x−4=32x−5x2+3x4g'(x)=\tfrac32x^{-1/2}-5x^{-2}+3x^{-4}=\frac{3}{2\sqrt x}-\frac{5}{x^2}+\frac{3}{x^4}

    Watch the last sign: (−1)(−3)=+3(-1)(-3)=+3. Domain: x\sqrt x needs x≥0x\ge0 and the reciprocals need x≠0x\ne0, so gg lives on x>0x>0 and the formula holds there. Check at x=1x=1: 32−5+3=−12\tfrac32-5+3=-\tfrac12.

  2. Q2easy

    Let h(x)=(x2+1)(x3−4x)h(x)=(x^2+1)(x^3-4x). Find h′(x)h'(x) with the product rule, then confirm it by expanding first.

    Hint

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

    Show answer

    Answer

    h′(x)=5x4−9x2−4h'(x)=5x^4-9x^2-4.

    Steps

    Product rule with f=x2+1f=x^2+1, g=x3−4xg=x^3-4x, f′=2xf'=2x, g′=3x2−4g'=3x^2-4:

    h′(x)=2x(x3−4x)+(x2+1)(3x2−4)=(2x4−8x2)+(3x4−x2−4)=5x4−9x2−4h'(x)=2x(x^3-4x)+(x^2+1)(3x^2-4)=(2x^4-8x^2)+(3x^4-x^2-4)=5x^4-9x^2-4

    Expanding first: h(x)=x5−3x3−4xh(x)=x^5-3x^3-4x, so h′(x)=5x4−9x2−4h'(x)=5x^4-9x^2-4. Same answer, and the degree dropped from 55 to 44 as it must.

  3. Q3medium

    ff and gg are differentiable with f(2)=3f(2)=3, f′(2)=−1f'(2)=-1, g(2)=−4g(2)=-4, g′(2)=5g'(2)=5. Find (a) (3f−2g)′(2)(3f-2g)'(2), (b) (fg)′(2)(fg)'(2), (c) (fg)′(2)\left(\frac fg\right)'(2), (d) (gf)′(2)\left(\frac gf\right)'(2), (e) ddx[x2f(x)]\frac{d}{dx}\big[x^2f(x)\big] at x=2x=2.

    Hint

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

    Show answer

    Answer

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

    Steps

    (a) 3f′(2)−2g′(2)=−3−10=−133f'(2)-2g'(2)=-3-10=-13.

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

    (c) f′(2)g(2)−f(2)g′(2)g(2)2=4−1516=−1116\dfrac{f'(2)g(2)-f(2)g'(2)}{g(2)^2}=\dfrac{4-15}{16}=-\dfrac{11}{16}.

    (d) g′(2)f(2)−g(2)f′(2)f(2)2=15−49=119\dfrac{g'(2)f(2)-g(2)f'(2)}{f(2)^2}=\dfrac{15-4}{9}=\dfrac{11}{9} — not the reciprocal of (c).

    (e) ddx[x2f(x)]=2xf(x)+x2f′(x)=2(2)(3)+4(−1)=8\dfrac{d}{dx}\big[x^2f(x)\big]=2xf(x)+x^2f'(x)=2(2)(3)+4(-1)=8.

  4. Q4medium

    Differentiate f(x)=x4−2x2+xx2f(x)=\dfrac{x^4-2x^2+\sqrt{x}}{x^2} without the quotient rule, and state the domain of f′f'.

    Hint

    A single power of xx underneath means divide term by term: xa/xb=xa−bx^a/x^b=x^{a-b}.

    Show answer

    Answer

    f′(x)=2x−32x5/2f'(x)=2x-\dfrac{3}{2x^{5/2}} for x>0x>0.

    Steps

    Divide term by term:

    f(x)=x4x2−2x2x2+x1/2x2=x2−2+x−3/2f(x)=\frac{x^4}{x^2}-\frac{2x^2}{x^2}+\frac{x^{1/2}}{x^2}=x^2-2+x^{-3/2}

    Domain: x\sqrt x needs x≥0x\ge0 and the division needs x≠0x\ne0, so x>0x>0.

    f′(x)=2x−32x−5/2=2x−32x5/2f'(x)=2x-\tfrac32x^{-5/2}=2x-\frac{3}{2x^{5/2}}

    Check at x=1x=1: f′(1)=2−32=12f'(1)=2-\tfrac32=\tfrac12.

  5. Q5medium

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

    Hint

    A fraction is zero only when its numerator is. Factor the numerator of f′f' and reject anything outside the domain.

    Show answer

    Answer

    (3,6)(3,6) and (−1,−2)(-1,-2).

    Steps

    Domain: x≠1x\ne1. Quotient rule with u=x2+3u=x^2+3, v=x−1v=x-1:

    f′(x)=2x(x−1)−(x2+3)(x−1)2=x2−2x−3(x−1)2=(x−3)(x+1)(x−1)2f'(x)=\frac{2x(x-1)-(x^2+3)}{(x-1)^2}=\frac{x^2-2x-3}{(x-1)^2}=\frac{(x-3)(x+1)}{(x-1)^2}

    f′(x)=0f'(x)=0 exactly when the numerator is zero: x=3x=3 or x=−1x=-1, both in the domain. Then f(3)=122=6f(3)=\frac{12}{2}=6 and f(−1)=4−2=−2f(-1)=\frac{4}{-2}=-2, so the tangent lines are y=6y=6 and y=−2y=-2.

  6. Q6hard

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

    Hint

    Write the numerator as x3/2−x1/2x^{3/2}-x^{1/2} before the quotient rule; clear x−1/2x^{-1/2} at the end by multiplying top and bottom by 2x2\sqrt x.

    Show answer

    Answer

    F′(x)=x2+4x−12x (x+1)2F'(x)=\dfrac{x^2+4x-1}{2\sqrt{x}\,(x+1)^2} for x>0x>0; horizontal tangent at x=−2+5≈0.236x=-2+\sqrt5\approx0.236.

    Steps

    FF is defined for x≥0x\ge0, but x\sqrt x has no derivative at 00, so work on x>0x>0. Numerator u=x3/2−x1/2u=x^{3/2}-x^{1/2}, u′=32x1/2−12x−1/2u'=\tfrac32x^{1/2}-\tfrac12x^{-1/2}; denominator v=x+1v=x+1, v′=1v'=1.

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

    Multiply top and bottom by 2x2\sqrt x:

    F′(x)=x2+4x−12x (x+1)2,x>0F'(x)=\frac{x^2+4x-1}{2\sqrt{x}\,(x+1)^2},\qquad x>0

    F′=0F'=0 when x2+4x−1=0x^2+4x-1=0, so x=−2±5x=-2\pm\sqrt5; only x=−2+5≈0.236x=-2+\sqrt5\approx0.236 lies in the domain.

06

On the exam

How this topic is marked

  • Markers want the rule set up before the algebra: name ff, gg, f′f', g′g' (or uu, vv), write the rule with those pieces in it, then simplify. A bare final answer with a slip earns nothing.

  • Never expand (x2+1)2(x^2+1)^2 in a denominator. Factor the numerator instead — the next part is almost always "where is the tangent horizontal?", and only the numerator can be zero.

  • Rewrite before you differentiate: roots and reciprocals as powers, one-term denominators divided through. It is faster, and it is where the sign and exponent errors live.

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