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)}}

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