009A Sample Final 2, Problem 3

Compute   ${\frac {dy}{dx}}.$

(a)  $y={\bigg (}{\frac {x^{2}+3}{x^{2}-1}}{\bigg )}^{3}$

(b)  $y=x\cos({\sqrt {x+1}})$

(c)  $y=\sin ^{-1}x$

Foundations:
1. Product Rule
${\frac {d}{dx}}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)$
2. Quotient Rule
${\frac {d}{dx}}{\bigg (}{\frac {f(x)}{g(x)}}{\bigg )}={\frac {g(x)f'(x)-f(x)g'(x)}{(g(x))^{2}}}$
3. Chain Rule
${\frac {d}{dx}}(f(g(x)))=f'(g(x))g'(x)$

Solution:

(a)

Step 1:
Using the Chain Rule, we have
${\frac {dy}{dx}}=3{\bigg (}{\frac {x^{2}+3}{x^{2}-1}}{\bigg )}^{2}{\bigg (}{\frac {x^{2}+3}{x^{2}-1}}{\bigg )}'.$

Step 2:
Now, using the Quotient Rule, we have
${\begin{array}{rcl}\displaystyle {\frac {dy}{dx}}&=&\displaystyle {3{\bigg (}{\frac {x^{2}+3}{x^{2}-1}}{\bigg )}^{2}{\bigg (}{\frac {x^{2}+3}{x^{2}-1}}{\bigg )}'}\\&&\\&=&\displaystyle {3{\bigg (}{\frac {x^{2}+3}{x^{2}-1}}{\bigg )}^{2}{\bigg (}{\frac {(x^{2}-1)(x^{2}+3)'-(x^{2}+3)(x^{2}-1)'}{(x^{2}-1)^{2}}}{\bigg )}}\\&&\\&=&\displaystyle {3{\bigg (}{\frac {x^{2}+3}{x^{2}-1}}{\bigg )}^{2}{\bigg (}{\frac {(x^{2}-1)(2x)-(x^{2}+3)(2x)}{(x^{2}-1)^{2}}}{\bigg )}}\\&&\\&=&\displaystyle {3{\bigg (}{\frac {x^{2}+3}{x^{2}-1}}{\bigg )}^{2}{\bigg (}{\frac {2x^{3}-2x-2x^{3}-6x}{(x^{2}-1)^{2}}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {3(x^{2}+3)^{2}(-8x)}{(x^{2}-1)^{4}}}.}\end{array}}$

(b)

Step 1:
Using the Product Rule, we have
${\frac {dy}{dx}}=x(\cos({\sqrt {x+1}}))'+(x)'\cos({\sqrt {x+1}}).$

Step 2:
Now, using the Chain Rule, we get
${\begin{array}{rcl}\displaystyle {\frac {dy}{dx}}&=&\displaystyle {x(\cos({\sqrt {x+1}}))'+(x)'\cos({\sqrt {x+1}})}\\&&\\&=&\displaystyle {x(-\sin({\sqrt {x+1}}))({\sqrt {x+1}})'+(1)\cos({\sqrt {x+1}})}\\&&\\&=&\displaystyle {-x\sin({\sqrt {x+1}}){\frac {1}{2{\sqrt {x+1}}}}(x+1)'+\cos({\sqrt {x+1}})}\\&&\\&=&\displaystyle {{\frac {-x\sin({\sqrt {x+1}})}{2{\sqrt {x+1}}}}+\cos({\sqrt {x+1}}).}\end{array}}$

(c)

Step 1:
Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=\sin(x)}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)=\sin ^{-1}x.}
These functions are inverses of each other 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 f(g(x))=x}   and   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(f(x))=x.}
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 \begin{array}{rcl} \displaystyle{g'(x)} & = & \displaystyle{\frac{1}{f'(g(x))}}\\ &&\\ & = & \displaystyle{\frac{1}{\cos(\sin^{-1} x)}.} \end{array}}
Now, let  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=\sin^{-1}(x).}   Then,  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 x=\sin(y).}
So,  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(\sin^{-1} x)=\cos(y).}
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 g'(x)=\frac{1}{\cos(y)}.}

Step 2:
Now, 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^2 y+\sin^2 y =1,}
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{\cos(y)} & = & \displaystyle{\sqrt{1-\sin^2 y}}\\ &&\\ & = & \displaystyle{\sqrt{1-x^2}.} \end{array}}
Hence,
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'(x)=\frac{1}{\sqrt{1-x^2}}.}

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{3(x^2+3)^2(-8x)}{(x^2-1)^4}}
(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{-x\sin(\sqrt{x+1})}{2\sqrt{x+1}}+\cos(\sqrt{x+1})}
(c)    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}{\sqrt{1-x^2}}}

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