009A Sample Final 1, Problem 3

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Find the derivatives of the following functions.

(a)   ${\displaystyle f(x)=\ln {\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}}$

(b)   ${\displaystyle g(x)=2\sin(4x)+4\tan({\sqrt {1+x^{3}}})}$

Foundations:
1. Chain Rule
${\displaystyle {\frac {d}{dx}}(f(g(x)))=f'(g(x))g'(x)}$
2. Quotient Rule
${\displaystyle {\frac {d}{dx}}{\bigg (}{\frac {f(x)}{g(x)}}{\bigg )}={\frac {g(x)f'(x)-f(x)g'(x)}{(g(x))^{2}}}}$
3. Trig Derivatives
${\displaystyle {\frac {d}{dx}}(\sin x)=\cos x,\quad {\frac {d}{dx}}(\tan x)=\sec ^{2}x}$

Solution:

(a)

Step 1:
Using the Chain Rule, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {{\frac {1}{{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}}}{\bigg (}{\frac {d}{dx}}{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {x^{2}+1}{x^{2}-1}}{\bigg (}{\frac {d}{dx}}{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}{\bigg )}.}\\\end{array}}}$

Step 2:
Now, we need to calculate  ${\displaystyle {\bigg (}{\frac {d}{dx}}{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}{\bigg )}.}$
To do this, we use the Quotient Rule. So, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {{\frac {x^{2}+1}{x^{2}-1}}{\bigg (}{\frac {d}{dx}}{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {x^{2}+1}{x^{2}-1}}{\bigg (}{\frac {(x^{2}+1)(2x)-(x^{2}-1)(2x)}{(x^{2}+1)^{2}}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {x^{2}+1}{x^{2}-1}}{\bigg (}{\frac {4x}{(x^{2}+1)^{2}}}{\bigg )}}\\&&\\&=&\displaystyle {\frac {4x}{(x^{2}-1)(x^{2}+1)}}\\&&\\&=&\displaystyle {{\frac {4x}{x^{4}-1}}.}\\\end{array}}}$

(b)

Step 1:
We need to use the Chain Rule. We have

${\displaystyle g'(x)\,=\,8\cos(4x)+4\sec ^{2}({\sqrt {1+x^{3}}}){\bigg (}{\frac {d}{dx}}{\sqrt {1+x^{3}}}{\bigg )}.}$

Step 2:
We need to calculate   ${\displaystyle {\frac {d}{dx}}{\sqrt {1+x^{3}}}.}$
We use the Chain Rule again to get

${\displaystyle {\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {8\cos(4x)+4\sec ^{2}({\sqrt {1+x^{3}}}){\bigg (}{\frac {d}{dx}}{\sqrt {1+x^{3}}}{\bigg )}}\\&&\\&=&\displaystyle {8\cos(4x)+4\sec ^{2}({\sqrt {1+x^{3}}}){\frac {1}{2}}(1+x^{3})^{-{\frac {1}{2}}}3x^{2}}\\&&\\&=&\displaystyle {8\cos(4x)+{\frac {6\sec ^{2}({\sqrt {1+x^{3}}})x^{2}}{\sqrt {1+x^{3}}}}.}\\\end{array}}}$

Final Answer:
(a)     ${\displaystyle f'(x)={\frac {4x}{x^{4}-1}}}$
(b)     ${\displaystyle g'(x)=8\cos(4x)+{\frac {6\sec ^{2}({\sqrt {1+x^{3}}})x^{2}}{\sqrt {1+x^{3}}}}}$