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 $f(x)=\sin(x)$ and $g(x)=\sin ^{-1}x.$
These functions are inverses of each other since
$f(g(x))=x$ and $g(f(x))=x.$
Therefore,
${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {\frac {1}{f'(g(x))}}\\&&\\&=&\displaystyle {{\frac {1}{\cos(\sin ^{-1}x)}}.}\end{array}}$
Now, let $y=\sin ^{-1}(x).$ Then, $x=\sin(y).$
So, $\cos(\sin ^{-1}x)=\cos(y).$
Therefore,
$g'(x)={\frac {1}{\cos(y)}}.$

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

Final Answer:
(a) ${\frac {3(x^{2}+3)^{2}(-8x)}{(x^{2}-1)^{4}}}$
(b) ${\frac {-x\sin({\sqrt {x+1}})}{2{\sqrt {x+1}}}}+\cos({\sqrt {x+1}})$
(c) ${\frac {1}{\sqrt {1-x^{2}}}}$

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