007A Sample Final 3, Problem 2 Detailed Solution

Find the derivative of the following functions:

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

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

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Background Information:
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}\cdot\frac{d}{d\theta}(\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}\cdot\frac{d}{d\theta}(\sec\theta -\sin 2\theta)}\\ &&\\ & = & \displaystyle{(-2)\pi^2(\sec\theta -\sin 2\theta)^{-3}\bigg(\sec\theta\tan\theta -\cos (2\theta)\cdot \frac{d}{d\theta} (2\theta)\bigg)}\\ &&\\ & = & \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'=\frac{d}{dx}(\cos(3\pi))+\frac{d}{dx}(\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 \frac{d}{dx}(\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'=\frac{d}{dx}(\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{\frac{d}{dx}(\tan^{-1}(\sqrt{x}))}\\ &&\\ & = & \displaystyle{\bigg(\frac{1}{1+(\sqrt{x})^2}\bigg)\cdot \frac{d}{dx}(\sqrt{x})}\\ &&\\ & = & \displaystyle{\bigg(\frac{1}{1+x}\bigg)\frac{1}{2\sqrt{x}}}\\ &&\\ & = & \displaystyle{\frac{1}{2\sqrt{x}(1+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 \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 \frac{1}{2\sqrt{x}(1+x)}}

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