# Math Insight

### Quotient rule and chain rule

Math 201, Spring 19
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Total points: 3
1. Calculate the derivative of $\displaystyle h(x)=\frac{\ln{\left (x \right )}}{1 - x}$ using the quotient rule by following these steps.
1. To write $\displaystyle h(x) = \frac{f(x)}{g(x)}$, the numerator must be $f(x) =$
and the denominator must be $g(x) =$
.
2. The quotient rule in terms of $f(x)$, $g(x)$, and their derivatives $\diff{f}{x}$, $\diff{g}{x}$ is
 $\displaystyle \diff{h}{x}=$ $-$
3. Now, we'll put it all together to calculate the derivative, plugging the functions from part (a) into the quotient rule formula of part (b). Since $\diff{f}{x}=$
and $\diff{g}{x} =$
the quotient rule for $\diff{h}{x}$ becomes
 $\displaystyle\diff{h}{x}=$ $\Bigl($ $\Bigr)-\Bigl($ $\Bigr)$

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

2. Calculate the following derivatives.
1. $\displaystyle h(x)=\frac{x^{2}}{1 - x^{2}}$
$h'(x) =$

$h'(3)=$
2. $\displaystyle g(x)=\frac{1 - e^{x}}{1 + e^{x}}$
$\displaystyle \diff{g}{x}=$

$\displaystyle \diff{g}{x}\bigg|_{x=0}=$

3. Use the following steps to compute the derivative of $h(x)=e^{2x}$.
1. Statement of the chain rule. The $f(x)$ be a differentiable function with derivative $f'(x)$ and $g(x)$ be a differentiable function with derivative $g'(x)$. Let $h(x)$ be their composition $h(x)=f(g(x))$. Then, the derivative of $h$ is the product of the derivatives of $f$ and $g$. The only trick is where to evaluate the derivatives $f'$ and $g'$. Fill in the blanks.
$h'(x) = f'\Bigl($
$\Bigr) g'($
$).$
2. If $g(x)=2 x$, then $g'(x)=$
. If $h(x)=f(g(x))=f(2x)$, what is $h'(x)$? Just take your formula from above and replace both $g(x)$ and $g'(x)$ with the known functions.
$h'(x) = f'\bigl($
$\bigr)$
=
$f'\bigl($
$\bigr).$
(After the second equal sign, just switch the order of the factors since it looks better that way.)
3. Let $g(x)=2 x$ as above. If $f(x)=e^{x}$, then $f'(x)=$
. If $h(x)=f(g(x))=$
, what is $h'(x)$? Just take your formula from above and replace $f'(\cdot)$ with the known function.
$h'(x) =$

4. Calculate the following derivatives with the chain rule.
1. Find $f'(x)$ for $f(x)=e^{x^{2}}$. Hint: let $g(x)=e^x$ and $h(x)=x^2$. Then, $f(x)=g(h(x))$ so $f'(x)=g'(h(x))h'(x)$.
$f'(x) =$
2. Find $\diff{g}{y}$ for $g(y)=e^{3 y^{2} - y}$. Hint: let $f(x)=e^x$ and $h(y)=3y^2-y$. Then $g(y)=f(h(y))$.
$\diff{g}{y}=$
3. Find $h'(z)$ for $h(z)=\ln{\left (z^{2} + 1 \right )}$. Hint: let $f(x)=\ln(x)$ and $g(z)=z^2+1$. Then $h(z)=f(g(z))$.
$h'(z) =$

5. Chain rule problems can be tricky. But, the only ones we'll deal with are those of the form $f(x)=e^{p(x)}$ or $f(x)=\ln(p(x))$ where $p(x)$ is a polynomial. Compute the derivative of the following functions.
1. $e^{5 x}$:
2. $9 e^{- \frac{x^{2}}{2}}$:
3. $e^{- x}$:
4. $\ln{\left (3 x^{2} - x \right )}$:
5. $\ln{\left (x^{5} - 2 x^{4} + 5 \right )}$:
6. $2 \ln{\left (x - 1 \right )}$:
7. $\ln{\left (\left(x - 1\right)^{2} \right )}=\ln{\left (x^{2} - 2 x + 1 \right )}$:
8. $e^{- z^{2}}$:
9. $2 e^{- 5 t}$:
10. $3 e^{- 5 t^{2}}$:
11. $\ln{\left (\frac{z}{5} \right )}$:
12. $\ln{\left (- 3 y \right )}$:

6. An important function to able to differentiate is a function of the form $f(t)=t^3 e^{-t/4}$. This requires both the chain rule and the product rule.
1. What is the derivative of $e^{- \frac{t}{4}}$?
Of $e^{- \frac{t}{5}}$?
Of $e^{- \frac{t}{6}}$?
Of $e^{- 2 t}$ ?
Of $e^{- 5 t}$ ?

(Practice such derivatives so you can calculate them quickly without realizing you are using the chain rule.)

2. What is the derivative of $t^{2}$?
Of $t^{5}$?
Of $3 t^{3}$?
3. Now, combine these two results with the product rule to calculate the derivative of $f(t)=t^{3} e^{- \frac{t}{4}}$. Write your final result as a polynomial times $e^{-t/4}$.
$f'(t) =$
4. Find the derivative of $g(t)=t e^{- 5 t}$.
$\diff{g}{t}=$
5. Find the derivative of $h(x)=3 x^{2} e^{- \frac{x}{3}}$.
$h'(x)=$