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,
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 y'=(\tan^{-1}(\sqrt{x}))'.}

Step 2:
Now, using the Chain Rule, we have
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{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)    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 g'(\theta)=\frac{-2\pi^2(\sec\theta\tan\theta -2\cos (2\theta))}{(\sec\theta -\sin 2\theta)^{3}}}
(b)    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 y'=\bigg(\frac{1}{1+x}\bigg)\frac{1}{2\sqrt{x}}}

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