007A Sample Final 1, Problem 3 Detailed Solution

Find the derivatives of the following functions.

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

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

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

Solution:

(a)

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

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

(b)

Step 1:
We need to use the Chain Rule. We have
${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {3\cos(4x)\cdot (4x)'+4\sec ^{2}({\sqrt {1+x^{3}}})\cdot {\frac {d}{dx}}({\sqrt {1+x^{3}}})}\\&&\\&=&\displaystyle {3\cos(4x)\cdot 4+4\sec ^{2}({\sqrt {1+x^{3}}})\cdot {\frac {d}{dx}}({\sqrt {1+x^{3}}})}\\&&\\&=&\displaystyle {12\cos(4x)+4\sec ^{2}({\sqrt {1+x^{3}}})\cdot {\frac {d}{dx}}({\sqrt {1+x^{3}}})}\\\end{array}}$

Step 2:
We need to calculate   ${\frac {d}{dx}}({\sqrt {1+x^{3}}}).$
We use the Chain Rule again to get
${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {12\cos(4x)+4\sec ^{2}({\sqrt {1+x^{3}}})\cdot {\frac {d}{dx}}{\sqrt {1+x^{3}}}}\\&&\\&=&\displaystyle {12\cos(4x)+4\sec ^{2}({\sqrt {1+x^{3}}}){\frac {1}{2}}(1+x^{3})^{-{\frac {1}{2}}}\cdot {\frac {d}{dx}}(1+x^{3})}\\&&\\&=&\displaystyle {12\cos(4x)+4\sec ^{2}({\sqrt {1+x^{3}}}){\frac {1}{2}}(1+x^{3})^{-{\frac {1}{2}}}(3x^{2})}\\&&\\&=&\displaystyle {12\cos(4x)+{\frac {6\sec ^{2}({\sqrt {1+x^{3}}})x^{2}}{\sqrt {1+x^{3}}}}.}\\\end{array}}$

Final Answer:
(a)     $f'(x)={\frac {4x}{x^{4}-1}}$
(b)     $g'(x)=12\cos(4x)+{\frac {6\sec ^{2}({\sqrt {1+x^{3}}})x^{2}}{\sqrt {1+x^{3}}}}$

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