# Help on derivative

• Jun 18th 2006, 10:33 AM
c_323_h
Help on derivative
1. $tan\frac{x}{y}=x+y$
2. $\sqrt{xy}=1+x^2y$
I know that you use implicit differentiation, but don't know when to use the chain rule or the product rule.

These are the answers in my book:

1. $y' = \frac{ysec^2(\frac{x}{y})-y^2}{y^2+xsec^2(\frac{x}{y})}$
2. $y' = \frac{4xy\sqrt{xy}-y}{x-2x^2\sqrt{xy}}$
• Jun 18th 2006, 10:48 AM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
1. $tan\frac{x}{y}=x+y$

Take derivative,
$(x/y)'\sec^2 (x/y)=1+y'$
Thus,
$\frac{(x)'y-x(y)'}{y^2}\cdot \sec^2(x/y)=1+y'$
Thus,
$\frac{y-xy'}{y^2}\cdot \sec^2(x/y)=1+y'$Thus,
$y-xy'=y^2\cos^2(x/y)+y^2\cos^2(x/y)y'$
Thus,
$y'(-x-y^2\cos^2(x/y))=y^2\cos^2(x/y)-y$
Thus,
$y'=\frac{y-y^2\cos^2(x/y)}{x+y^2\cos^2(x/y)}$
• Jun 21st 2006, 05:58 AM
topsquark
Quote:

Originally Posted by c_323_h
2. $\sqrt{xy}=1+x^2y$

$\sqrt{xy}=1+x^2y$

$\frac{1}{2}\frac{1}{\sqrt{xy}}(y+xy')=2xy'$

$\frac{1}{2}\frac{1}{\sqrt{xy}}xy'-2xy'=-\frac{1}{2}\frac{1}{\sqrt{xy}}y$

$\left ( \frac{1}{2}\sqrt{\frac{x}{y}} - 2x \right ) y' = -\frac{1}{2}\sqrt{\frac{y}{x}}$

$y' = -\frac{\frac{1}{2}\sqrt{\frac{y}{x}}}{\frac{1}{2} \sqrt{\frac{x}{y}} - 2x}$

$y' = -\frac{\frac{1}{2}\sqrt{\frac{y}{x}}}{\frac{1}{2} \sqrt{\frac{x}{y}} - 2x}*\frac{2\sqrt{xy}}{2\sqrt{xy}}$ = $\frac{y}{x-4x \sqrt{xy}}=\frac{y}{x}\frac{1}{1-4 \sqrt{xy}}$

-Dan