009A Sample Final 1, Problem 3

Find the derivatives of the following functions.

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

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

Foundations:
Review chain rule, quotient rule, and derivatives of trig functions

Solution:

(a)

Step 1:
Using the chain rule, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {{\frac {1}{{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}}}{\bigg (}{\frac {d}{dx}}{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {x^{2}+1}{x^{2}-1}}{\bigg (}{\frac {d}{dx}}{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}{\bigg )}}\\\end{array}}$

Step 2:
Now, we need to calculate ${\bigg (}{\frac {d}{dx}}{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}{\bigg )}$.
To do this, we use the Chain Rule. So, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {{\frac {x^{2}+1}{x^{2}-1}}{\bigg (}{\frac {d}{dx}}{\bigg (}{\frac {x^{2}-1}{x^{2}+1}}{\bigg )}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {x^{2}+1}{x^{2}-1}}{\bigg (}{\frac {(x^{2}+1)(2x)-(x^{2}-1)(2x)}{(x^{2}+1)^{2}}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {x^{2}+1}{x^{2}-1}}{\bigg (}{\frac {4x}{(x^{2}+1)^{2}}}{\bigg )}}\\&&\\&=&\displaystyle {\frac {4x}{(x^{2}-1)(x^{2}+1)}}\\&&\\&=&\displaystyle {\frac {4x}{x^{4}-1}}\\\end{array}}$

(b)

Step 1:

Step 2:

Step 3:

Final Answer:
(a) $f'(x)={\frac {4x}{x^{4}-1}}$
(b)

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