009A Sample Final 3, Problem 2

Find the derivative of the following functions:

(a)  $g(\theta )={\frac {\pi ^{2}}{(\sec \theta -\sin 2\theta )^{2}}}$

(b)  $y=\cos(3\pi )+\tan ^{-1}({\sqrt {x}})$

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

Solution:

(a)

Step 1:
First, we write
$g(\theta )=\pi ^{2}(\sec \theta -\sin 2\theta )^{-2}.$
Now, using the Chain Rule, we have
$g'(\theta )=(-2)\pi ^{2}(\sec \theta -\sin 2\theta )^{-3}(\sec \theta -\sin 2\theta )'.$

Step 2:
Now, using the Chain Rule a second time, we get
${\begin{array}{rcl}\displaystyle {g'(\theta )}&=&\displaystyle {(-2)\pi ^{2}(\sec \theta -\sin 2\theta )^{-3}(\sec \theta -\sin 2\theta )'}\\&&\\&=&\displaystyle {(-2)\pi ^{2}(\sec \theta -\sin 2\theta )^{-3}(\sec \theta \tan \theta -\cos(2\theta )(2\theta )')}\\&&\\&=&\displaystyle {(-2)\pi ^{2}(\sec \theta -\sin 2\theta )^{-3}(\sec \theta \tan \theta -\cos(2\theta )(2))}\\&&\\&=&\displaystyle {{\frac {-2\pi ^{2}(\sec \theta \tan \theta -2\cos(2\theta ))}{(\sec \theta -\sin 2\theta )^{3}}}.}\end{array}}$

(b)

Step 1:
First, we have
$y'=(\cos(3\pi ))'+(\tan ^{-1}({\sqrt {x}}))'.$
Since  $\cos(3\pi )$  is a constant,
we have
$(\cos(3\pi ))'=0.$
Therefore,
$y'=(\tan ^{-1}({\sqrt {x}}))'.$

Step 2:
Now, using the Chain Rule, we have
${\begin{array}{rcl}\displaystyle {y'}&=&\displaystyle {(\tan ^{-1}({\sqrt {x}}))'}\\&&\\&=&\displaystyle {{\bigg (}{\frac {1}{1+({\sqrt {x}})^{2}}}{\bigg )}({\sqrt {x}})'}\\&&\\&=&\displaystyle {{\bigg (}{\frac {1}{1+x}}{\bigg )}{\frac {1}{2{\sqrt {x}}}}.}\end{array}}$

Final Answer:
(a)    $g'(\theta )={\frac {-2\pi ^{2}(\sec \theta \tan \theta -2\cos(2\theta ))}{(\sec \theta -\sin 2\theta )^{3}}}$
(b)    $y'={\bigg (}{\frac {1}{1+x}}{\bigg )}{\frac {1}{2{\sqrt {x}}}}$

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