Find the derivative of the following functions:
(a) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(\theta )={\frac {\pi ^{2}}{(\sec \theta -\sin 2\theta )^{2}}}}
(b) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y=\cos(3\pi )+\tan ^{-1}({\sqrt {x}})}
| Background Information:
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| 1. Chain Rule
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| 2. Trig Derivatives
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| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d}{dx}}(\sin x)=\cos x,\quad {\frac {d}{dx}}(\sec x)=\sec x\tan x}
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| 3. Inverse Trig Derivatives
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| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d}{dx}}(\tan ^{-1}x)={\frac {1}{1+x^{2}}}}
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Solution:
(a)
| Step 1:
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| First, we write
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| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(\theta )=\pi ^{2}(\sec \theta -\sin 2\theta )^{-2}.}
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| Now, using the Chain Rule, we have
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| 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).}
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| Step 2:
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| Now, using the Chain Rule a second time, we get
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| 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}}
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(b)
| Step 1:
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| First, we have
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| 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})).}
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| 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,
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| we have
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| 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.}
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| Therefore,
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| 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})).}
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| Step 2:
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| Now, using the Chain Rule, we have
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| 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}}
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| Final Answer:
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| (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}}}
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| (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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