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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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}}