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

Solution:

(a)

Step 1:
First, we write
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)=\pi^2(\sec\theta -\sin 2\theta)^{-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 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
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'(\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
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'=(\cos(3\pi))'+(\tan^{-1}(\sqrt{x}))'.}
Since  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 \cos(3\pi)}   is a constant,
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 (\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)    $y'={\bigg (}{\frac {1}{1+x}}{\bigg )}{\frac {1}{2{\sqrt {x}}}}$

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