009A Sample Final 3, Problem 2 Detailed Solution
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Find the derivative of the following functions:
(a)
(b)
| Background Information: | |
|---|---|
| 1. Chain Rule | |
| 2. Trig Derivatives | |
| 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} | |
| 3. Inverse Trig Derivatives | |
| 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}}}} |
Solution:
(a)
| Step 1: |
|---|
| First, we write |
| 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}.} |
| Now, using the Chain Rule, we have |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y'={\frac {d}{dx}}(\cos(3\pi ))+{\frac {d}{dx}}(\tan ^{-1}({\sqrt {x}})).} |
| Since is a constant, |
| we have |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d}{dx}}(\cos(3\pi ))=0.} |
| Therefore, |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle y'={\frac {d}{dx}}(\tan ^{-1}({\sqrt {x}})).} |
| Step 2: |
|---|
| Now, using the Chain Rule, we have |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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)}} |