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