4.1 Arithmetic of derivatives

Look at the Sum rule

True

True: this is exactly what the Sum Rule states.

Try an example, like $f(x)=g(x)=x$.

False, in general

False, in general. The product rule tells us $\diff{}{x}\{f(x)g(x)\}=f'(x)g(x)+f(x)g'(x)$. An easy example of why we can't do it the other way is to take $f(x)=g(x)=x$. Then the equation becomes $\diff{}{x}\{x^2\}=(1)(1)$, which is false.

Simplify

True

True: the quotient rule tells us

$g(x)=f(x)+f(x)+f(x)$

If you're creative, you can find lots of ways to differentiate!
Constant multiple: $g'(x)=3f'(x)$.
Product rule: $g'(x) = \diff{}{x}\{3\}f(x)+3f'(x)=0f(x)+3f'(x)=3f'(x)$.
Sum rule: $g'(x)=\diff{}{x}\{f(x)+f(x)+f(x)\}=f'(x)+f'(x)+f'(x)=3f'(x)$.
Quotient rule: $g'(x)=\diff{}{x}\left{\frac{f(x)}{\frac{1}{3}}\right}= \frac{\frac{1}{3}f'(x)-f(x)(0)}{\frac{1}{9}}=\frac{\frac{1}{3}f'(x)}{\frac{1}{9}}=9\left(\frac{1}{3}\right)f'(x)=3f'(x)$.
All rules give $g'(x)=3f'(x)$.

If you're creative, you can find lots of ways to differentiate!
Constant multiple: $g'(x)=3f'(x)$.
Product rule: $g'(x) = \diff{}{x}\{3\}f(x)+3f'(x)=0f(x)+3f'(x)=3f'(x)$.
Sum rule: $g'(x)=\diff{}{x}\{f(x)+f(x)+f(x)\}=f'(x)+f'(x)+f'(x)=3f'(x)$.
Quotient rule: $g'(x)=\diff{}{x}\left{\frac{f(x)}{\frac{1}{3}}\right}= \frac{\frac{1}{3}f'(x)-f(x)(0)}{\frac{1}{9}}=\frac{\frac{1}{3}f'(x)}{\frac{1}{9}}=9\left(\frac{1}{3}\right)f'(x)=3f'(x)$.
All rules give $g'(x)=3f'(x)$.

Use linearity and the known derivatives of $x^2$ and $x^{1/2}$.

$f'(x)=6x+\frac{2}{\sqrt{x}}$

We know, from Examples 3.3.10 and 3.3.15 in the text, that $\diff{}{x}x^2=2x$ and $\diff{}{x}x^{1/2}=\frac{1}{2\sqrt{x}}$. So, by linearity,

Remember $\ds\diff{}{x}\{2^x\}=2^x\log2$.

$f^{(n)}=2^x(\log 2)^n$

We differentiate a few time to find the pattern.

Every time we differentiate, we multiply the original function by another factor of $\log 2$. So, the $n$th derivative is given by:

You have already seen $\diff{}{x}\{\sqrt{x}\}$.

$-36x+24\sqrt{x}+\frac{20}{\sqrt{x}}-45$

We have already seen $\diff{}{x}\{\sqrt{x}\}=\frac{1}{2\sqrt{x}}$ in Example 3.3.15 of the text. Now:

The equation of a line can be determined using a point, and the slope. The derivative of $x^3$ can be found by writing $x^3=(x)(x^2)$.

$y - \frac{1}{8} = \frac{3}{4}\cdot \left(x-\frac{1}{2}\right)$, or $y= \tfrac{3}{4} x - \tfrac{1}{4}$

We already know that $\diff{}{x}x=1$ and $\diff{}{x}x^2=2x$, so we can compute the derivative of $x^3$ by writing $x^3=(x)(x^2)$,

When this is evaluated at $x=\frac{1}{2}$ we get $\frac{3}{4}$. Since we also compute $\left( \frac{1}{2}\right)^3=\frac{1}{8}$, the equation of the tangent line is

Be careful to distinguish between speed and velocity.

(a) $4$ (b) left (c) decreasing

Let $f(t)=t^3-4t^2+1$. We saw in Question 8 that $\diff{}{t}t^3=3t^2$. So

Hence at $t=2$, (a) the particle has speed of magnitude 4, and (b) is moving towards the left. At $t=2$, $f''(2)>0$, so $f'$ is increasing, i.e. becoming less negative. Since $f'$ is getting closer to zero, (c) the magnitude of the speed is decreasing.

$\dfrac{1}{{(x+1/2)}^2}$, or $\dfrac{4}{(2x+1)^2}$

We can use the quotient rule here.

How do you take care of that power?

$-72$

First, we find the $y'$ for general $x$. Using the corollary to Theorem 4.1.3 and the quotient rule:

So, plugging in $x=1$:

After you differentiate, factor out $e^x$.

$g'(x)=[f(x)+f'(x)]e^x$

Using the product rule, $g'(x)=f'(x)e^x+f(x)e^x=[f(x)+f'(x)]e^x$

Population growth is rate of change of population.

$b'(t)-d'(t)$

Population growth is rate of change of population. Population in year $2000+t$ is given by $P(t)=P_0+b(t)-d(t)$, where $P_0$ is the initial population of the town. Then $P'(t)$ is the expression we're looking for, and $P'(t)=b'(t)-d'(t)$.

It is interesting to note that the initial population does not obviously show up in this calculation. It would probably affect $b(t)$ and $d(t)$, but if we know these we do not need to know $P_0$ to answer our question.

$(1,3),\ (3,27)$

We already know that $\diff{}{x}x^2=2x$. So the slope of $y=3x^2$ at $x=a$ is $6a$. The tangent line to $y=3x^2$ at $x=a, y=3a^2$ is $y-3a^2=6a(x-a)$. This tangent line passes through $(2,9)$ if

The points are $(1,3),\ (3,27)$.

Interpret it as a derivative that you know how to compute.

$\dfrac{1}{2\sqrt{100180}}$

This limit represents the derivative computed at $x=100180$ of the function $f(x)=\sqrt{x}$. Since the derivative of $f(x)$ is $\dfrac{1}{2\sqrt{x}}$, then its value at $x=100180$ is exactly $\dfrac{1}{2\sqrt{100180}}$.

The answer is not 10 square metres per second.

$20t+7$ square metres per second.

Let $w(t)$ and $l(t)$ be the width and length of the rectangle. Given in the problem is that $w'(t)=2$ and $l'(t)=5$. Since both functions have constant slopes, both must be lines. Their slopes are given, and their intercepts are $w(0)=l(0)=1$. So, $w(t)=2t+1$ and $l(t)=5t+1$.

The area of the rectangle is $A(t)=w(t)\cdot l(t)$, so using the product rule, the rate at which the area is increasing is $A'(t)=w'(t)l(t)+w(t)l'(t)=2(5t+1)+5(2t+1)=20t+7$ square metres per second.

You don't need to know $g(0)$ or $g'(0)$.

0

Using the product rule, $f'(x)=(2x)g(x)+x^2g'(x)$, so $f'(0)=0\cdot g(x)+0\cdot g'(x)=0$. (Since $g$ is differentiable, $g'$ exists.)

First expression, $f(x)=\dfrac{g(x)}{h(x)}$:

Second expresson, $f(x)=\dfrac{g(x)}{k(x)}\cdot\dfrac{k(x)}{h(x)}$:

and this is exactly what we got from differentiating the first expression.

First expression, $f(x)=\dfrac{g(x)}{h(x)}$:

Second expresson, $f(x)=\dfrac{g(x)}{k(x)}\cdot\dfrac{k(x)}{h(x)}$:

and this is exactly what we got from differentiating the first expression.

In order to be differentiable, a function should be continuous. To determine the differentiability of the function at $x=1$, use the definition of the derivative.

$a=b=\dfrac{e}{2}$

When we say a function is differentiable without specifying a range, we mean that it is differentiable over its domain. The function $f(x)$ is differentiable when $x \neq 1$ for any values of $a$ and $b$; it is up to us to figure out which constants make it differentiable when $x=1$.

In order to be differentiable, a function must be continuous. The definition of continuity tells us that, for $f$ to be continuous at $x=1$, we need $\ds\lim_{x \to 1}f(x)=f(1)$. From the definition of $f$, we see $f(1)=a+b=\ds\lim_{x \to 1^-}f(x)$, so we need $\ds\lim_{x\to 1^+}f(x)=a+b$. Since $\ds\lim_{x \to 1^+}f(x)=e^1=e$, we specifically need

Now, let's consider differentiability of $f$ at $x=1$. We need the following limit to exist:

In particular, we need the one-sided limits to exist and be equal:

If $h<0$, then $1+h<1$, so $f(1+h)=a(1+h)^2+b$. If $h>0$, then $1+h>1$, so $f(1+h)=e^{1+h}$. With this in mind, we begin to evaluate the one-sided limits:

Since we take $a+b$ to be equal to $e$ (to ensure continuity):

So, we also need

Therefore, the values of $a$ and $b$ that make $f$ differentiable are $a=b=\dfrac{e}{2}$.

Review Pascal's Triangle.

(a) $g''(x)=[f(x)+2f'(x)+f''(x)]e^x$
(b) $g'''(x)=[f(x)+3f'(x)+3f''(x)+f'''(x)]e^x$
(c) $g^{(4)}(x)=[f(x)+4f'(x)+6f''(x)+4f'''(x)+f^{(4)}(x)]e^x$

(a) Using the product rule,

(b) Using the product rule and our answer from (a),

(c) We notice that the coefficients of the derivatives of $f$ correspond to the entries in the rows of Pascal's Triangle.

That is, we guess

This is verified by differentiating our answer from (a) using the product rule:

Check signs

In the quotient rule, there is a minus, not a plus. Also, $2(x+1)+2x$ is not the same as $2(x+1)$.

The correct version is:

In the quotient rule, there is a minus, not a plus. Also, $2(x+1)+2x$ is not the same as $2(x+1)$.

The correct version is:

Read Lemma 4.1.14 carefully.

False

False: Lemma 4.1.14 tells us that, for a constant $n$, $\ds\diff{}{x}\{x^n\}=nx^{n-1}$. Note that the base $x$ is the variable and the exponent $n$ is a constant. In the equation given in the question, the base $2$ is a constant, and the exponent $x$ is the variable: this is the opposite of the situation where Lemma 4.1.14 applies.

We do not yet know how to differentiate $2^x$. We'll learn about it in Section 3.5.

Quotient rule

$\dfrac{(x-1)e^x}{2x^2}$

Using the quotient rule,

$e^{2x}=\left(e^x\right)^2$

$2e^{2x}$

$e^{a+x}=e^ae^x$

$e^{a+x}$