# 009A Sample Midterm 2, Problem 5 Detailed Solution

Find the derivatives of the following functions. Do not simplify.

(a)   ${\displaystyle f(x)=\tan ^{3}(7x^{2}+5)}$

(b)   ${\displaystyle g(x)=\sin(\cos(e^{x}))}$

(c)   ${\displaystyle h(x)={\frac {(5x^{2}+7x)^{3}}{\ln(x^{2}+1)}}}$

Background Information:
1. Chain Rule
${\displaystyle {\frac {d}{dx}}(f(g(x)))=f'(g(x))g'(x)}$
2. Trig Derivatives
${\displaystyle {\frac {d}{dx}}(\sin x)=\cos x,\quad {\frac {d}{dx}}(\cos x)=-\sin x}$
3. Quotient Rule
${\displaystyle {\frac {d}{dx}}{\bigg (}{\frac {f(x)}{g(x)}}{\bigg )}={\frac {g(x)f'(x)-f(x)g'(x)}{(g(x))^{2}}}}$
4. Derivative of natural logarithm
${\displaystyle {\frac {d}{dx}}(\ln x)={\frac {1}{x}}}$

Solution:

(a)

Step 1:
First, we use the Chain Rule to get
${\displaystyle f'(x)=3\tan ^{2}(7x^{2}+5)(\tan(7x^{2}+5))'.}$
Step 2:
Now, we use the Chain Rule again to get

${\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {3\tan ^{2}(7x^{2}+5)(\tan(7x^{2}+5))'}\\&&\\&=&\displaystyle {3\tan ^{2}(7x^{2}+5)\sec ^{2}(7x^{2}+5)(7x^{2}+5)'}\\&&\\&=&\displaystyle {3\tan ^{2}(7x^{2}+5)\sec ^{2}(7x^{2}+5)(14x).}\end{array}}}$

(b)

Step 1:
First, we use the Chain Rule to get
${\displaystyle g'(x)=\cos(\cos(e^{x}))(\cos(e^{x}))'.}$
Step 2:
Now, we use the Chain Rule again to get

${\displaystyle {\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {\cos(\cos(e^{x}))(\cos(e^{x}))'}\\&&\\&=&\displaystyle {\cos(\cos(e^{x}))(-\sin(e^{x}))(e^{x})'}\\&&\\&=&\displaystyle {\cos(\cos(e^{x}))(-\sin(e^{x}))(e^{x}).}\end{array}}}$

(c)

Step 1:
First, we use the Quotient Rule to get
${\displaystyle h'(x)={\frac {\ln(x^{2}+1)((5x^{2}+7x)^{3})'-(5x^{2}+7x)^{3}(\ln(x^{2}+1))'}{(\ln(x^{2}+1))^{2}}}.}$
Step 2:
Now, we use the Chain Rule to get
${\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {\frac {\ln(x^{2}+1)((5x^{2}+7x)^{3})'-(5x^{2}+7x)^{3}(\ln(x^{2}+1))'}{(\ln(x^{2}+1))^{2}}}\\&&\\&=&\displaystyle {\frac {\ln(x^{2}+1)3(5x^{2}+7x)^{2}(5x^{2}+7x)'-(5x^{2}+7x)^{3}{\frac {1}{x^{2}+1}}(x^{2}+1)'}{(\ln(x^{2}+1))^{2}}}\\&&\\&=&\displaystyle {{\frac {\ln(x^{2}+1)3(5x^{2}+7x)^{2}(10x+7)-(5x^{2}+7x)^{3}{\frac {1}{x^{2}+1}}(2x)}{(\ln(x^{2}+1))^{2}}}.}\end{array}}}$

(a)     ${\displaystyle f'(x)=3\tan ^{2}(7x^{2}+5)\sec ^{2}(7x^{2}+5)(14x)}$
(b)     ${\displaystyle g'(x)=\cos(\cos(e^{x}))(-\sin(e^{x}))(e^{x})}$
(c)     ${\displaystyle h'(x)={\frac {\ln(x^{2}+1)3(5x^{2}+7x)^{2}(10x+7)-(5x^{2}+7x)^{3}{\frac {1}{x^{2}+1}}(2x)}{(\ln(x^{2}+1))^{2}}}}$