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

1

Foundations:
For functions $f(x)$ and $g(x)$ , recall

Chain Rule: ${\frac {d}{dx}}(f(g(x)))=f'(g(x))g'(x)$

Quotient Rule: ${\frac {d}{dx}}{\bigg (}{\frac {f(x)}{g(x)}}{\bigg )}={\frac {g(x)f'(x)-f(x)g'(x)}{(g(x))^{2}}}$

Trig Derivatives: ${\frac {d}{dx}}(\sin x)=\cos x,\quad {\frac {d}{dx}}(\tan x)=\sec ^{2}x$

Solution:

(a)

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

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 Quotient 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:
Again, we need to use the Chain Rule. We have
$g'(x)=8\cos(4x)+4\sec ^{2}({\sqrt {1+x^{3}}}){\bigg (}{\frac {d}{dx}}{\sqrt {1+x^{3}}}{\bigg )}.$

Step 2:
We need to calculate ${\frac {d}{dx}}{\sqrt {1+x^{3}}}.$
We use the Chain Rule again to get
${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {8\cos(4x)+4\sec ^{2}({\sqrt {1+x^{3}}}){\bigg (}{\frac {d}{dx}}{\sqrt {1+x^{3}}}{\bigg )}}\\&&\\&=&\displaystyle {8\cos(4x)+4\sec ^{2}({\sqrt {1+x^{3}}}){\frac {1}{2}}(1+x^{3})^{-{\frac {1}{2}}}3x^{2}}\\&&\\&=&\displaystyle {8\cos(4x)+{\frac {6\sec ^{2}({\sqrt {1+x^{3}}})x^{2}}{\sqrt {1+x^{3}}}}.}\\\end{array}}$

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