# 009A Sample Final 2, Problem 3

Compute   ${\displaystyle {\frac {dy}{dx}}.}$

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

(b)  ${\displaystyle y=x\cos({\sqrt {x+1}})}$

(c)  ${\displaystyle y=\sin ^{-1}x}$

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

Solution:

(a)

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

(b)

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

(c)

Step 1:
Let  ${\displaystyle y=\sin ^{-1}(x).}$  Then,
${\displaystyle \sin(y)=x}$
for  ${\displaystyle y}$  in the interval  ${\displaystyle {\bigg [}-{\frac {\pi }{2}},{\frac {\pi }{2}}{\bigg ]}.}$
Using implicit differentiation, we have
${\displaystyle \cos(y){\frac {dy}{dx}}=1.}$
Solving for  ${\displaystyle {\frac {dy}{dx}},}$  we get
${\displaystyle {\frac {dy}{dx}}={\frac {1}{\cos(y)}}.}$
Step 2:
Now, since  ${\displaystyle \sin(y)=x,}$  we have the following diagram.
(Insert diagram)
Therefore,
${\displaystyle \cos(y)={\sqrt {1-x^{2}}}.}$
Hence,
${\displaystyle {\begin{array}{rcl}\displaystyle {\frac {dy}{dx}}&=&\displaystyle {\frac {1}{\cos(y)}}\\&&\\&=&\displaystyle {{\frac {1}{\sqrt {1-x^{2}}}}.}\end{array}}}$

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