# Math Insight

### Derivative of exponentials, logarithms, and products

Math 201, Spring 22
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Total points: 3
1. Imagine that $f(x)$ is a function whose derivative is the function itself. The statement “the derivative is the function itself” means that $f'(x)=$
.

The exponential function $f(x)=e^x$, where $e \approx 2.71828$, is the one such function. Using the above formula, if $f(x)=e^x$, then it must be that $f'(x)=$
. In particular, $f'(0)=$
.

If $g(z)=17 e^{z} - 9$, then
$\displaystyle\diff{g}{z} =$

$\displaystyle\diff{g}{z} \biggr|_{z=1} =$

If $h(y)=5 e^{y} - y^{2}$, then
$h'(y) =$

$h'(-3) =$

2. The natural log function $\ln(x)$ is the inverse of the exponential function $e^x$, meaning that $\ln(e^x)=x$ and $e^{\ln(x)}=x$.
1. If $f(x)=\ln(x)$, then $f'(x)=$
and $f'(5) =$
.
2. Calculate $\diff{}{x} \left(5 \ln{\left (x \right )} + 3\right)=$
3. We can use the logarithm rules to differentiate $g(y)=\ln(y^3)$. First of all, according to the rules, you can bring down the exponent to rewrite
$g(y)=\ln(y^3) =$
.
In that new form, $g$ can be differentiated just as above.

Calculate $g'(y)=$
and $g'(3) =$
.

If $h(t)=\ln(t^c)$ for a constant $c$, calculate $h'(t)=$
.

4. We can use the logarithm rules to differentiate $p(x)=\ln(6x)$. First of all, according to the rules, write the logarithm of a product as the sum of the logarithms:
$p(x)=\ln(6x) =$
.
In that new form, $p$ can be differentiated just as above.

Calculate $p'(x)=$
and $p'(5) =$
.

If $q(x)=\ln(cx)$ for a constant $c$, calculate $q'(x)=$
.

5. $\diff{}{x}\left(6 \ln{\left (x \right )} + 4 \ln{\left (x^{2} \right )} + 5\right) =$
$=$

(In the first blank, you can write any expression that simplifies to the correct answer. In the second blank, be sure to simplify your answer all the way.)
6. $\diff{}{t}\left(\ln{\left (5 t \right )} - \ln{\left (t^{3} \right )}\right) =$
$=$

(In the first blank, you can write any expression that simplifies to the correct answer. In the second blank, be sure to simplify your answer all the way.)

3. The product rule is an extension of the constant product rule, which you used to differentiate polynomials.
1. Constant product rule. Let $f(x)$ be a differentiable function with derivative $\diff{f}{x}$. Let $g(x)=a f(x)$ and $h(x)=f(x)b$ for some constant numbers $a$ and $b$. Compute:
$\diff{g}{x} =$

$\diff{h}{x} =$
2. Product rule. If $f(x)$ is a differentiable function with derivative $\diff{f}{x}$ and $g(x)$ is a differentiable function with derivative $\diff{g}{x}$, what is the derivative of the product $h(x)=f(x)g(x)$?
$\diff{h}{x}=$
3. Constant product is a special case. Let $g(x)=c$ for a constant $c$, then $\diff{g}{x} =$
. Let $h(x)=f(x)g(x)$ for differentiable function $f(x)$ with derivative $\diff{f}{x}$. Calculate $\diff{h}{x}$ using the product rule. Then, substitute the computed value for $\diff{g}{x}$ for the special case $g(x)=c$. You should obtain the constant product rule.

$\diff{h}{x}=$
$=$

(In the first blank, enter the product rule, as in the previous part; in the second blank enter the constant product rule, i.e., the rule specialized for the case $g(x)=c$.)

4. Let $f(x)=x^2$, let $g(x)=x^3$, and let $h(x)=f(x)g(x)=x^5$. Calculate the following derivatives. (For $\diff{h}{x}$ don't use the product rule, just differentiate $x^5$ directly.)
$\diff{f}{x} =$

$\diff{g}{x} =$

$\diff{h}{x} =$

Now verify you get the same answer with the product rule:
$\diff{h}{x} = \diff{f}{x}g(x) + f(x)\diff{g}{x}$
$= ($
$)($
$)+($
$)($
$)$
$=$
$+$

$=$

(In second line, enter values for functions and derivatives computed above. Then, perform the multiplications in third line.)

4. Calculate the following derivatives using the product rule.
1. For $f(x)=x e^{x}$, calculate $f'(x)=$

$f'(0) =$
2. If $g(x)=\left(x^{2} - x\right) \ln{\left (x \right )}$, then $\diff{g}{x} =$

Calculate $\diff{g}{x}\biggr|_{x=1} =$
.
3. Find $\diff{}{x} \left(e^{x} \ln{\left (x \right )} \right)=$
.
4. Let $h(z)=z^{2}$ and let $g(z)=c + e^{z}$ for some number $c$. Find $f'(z)$, where $f(z)=h(z)g(z)$.

$f'(z)=$

5. Calculate $\diff{}{t} \left(\left(c + t\right) \ln{\left (t \right )} \right)=$

5. Cells in a tissue are growing and dividing. As a result, the volume of the tissue is increasing exponentially, leading to a volume $V(t)=0.1 e^{t}$ mm3 after $t$ hours. At the same time, the cells pack in more densely, leading to a linear increase of the density of the tissue so that the density after $t$ hours is $\rho(t) = 0.01 t + 0.1$ mg/mm3.
1. What is the mass of the tissue as a function of time?
Mass =
=
mg
(In the first blank, enter an answer in terms of V and $\rho$. In the second blank, write a formula in terms of $t$, that gives a result in mg.)
2. What is the rate of change of the volume?
mm3/hour.
3. What is the rate of change of the density?
mg/mm3/hour.
4. What is the rate of change of the mass?
mg/hour

6. Follow these steps to use the product rule to compute the derivative of $f(t)=e^{2t}$.

First, thinking of $2t=t+t$, use the rules of exponentiation to convert $e^{t+t}$ into a product of two exponentials. $f(t) =$
$\cdot$

Each of those two exponentials in the above formula should be simple to differentiate. The product rule now makes it possible to calculate $f'(t)$.
$f'(t) =$
$\cdot$
$+$
$\cdot$

To simplify the formula for $f'(t)$, use the same rule of exponentiation in the other direction, combine each term in formula for $f'(t)$ into a single exponential.
$f'(t) =$
$+$
.

Lastly, combine the terms to get a simple formula for $f'(t)$.
$f'(t) =$