# [SOLVED] differentiation...

• Jul 10th 2009, 01:11 AM
darachit
[SOLVED] differentiation...
i am stuck in the end...one term does not match. otherwise the answer wud come...
question is:
if $\displaystyle x^p*y^q=(x+y)^{p+q}$ then prove that dy/dx =y/x and d/dx(dy/dx)=0.
• Jul 10th 2009, 01:51 AM
Hello darachit
Quote:

Originally Posted by darachit
i am stuck in the end...one term does not match. otherwise the answer wud come...
question is:
if $\displaystyle x^p*y^q=(x+y)^{p+q}$ then prove that dy/dx =y/x and d/dx(dy/dx)=0.

Take logs of both sides:

$\displaystyle p\ln x + q \ln y = (p+q)\ln(x+y)$

$\displaystyle \Rightarrow \frac{p}{x}+ \frac{q}{y}\frac{dy}{dx}=\frac{p+q}{x+y}\left(1+\f rac{dy}{dx}\right)$

If you now solve for $\displaystyle \frac{dy}{dx}$ and simplify, the equation becomes

$\displaystyle \frac{dy}{dx}\Big(x(qx-py)\Big)= y(qx-py)$

$\displaystyle \Rightarrow \frac{dy}{dx}=\frac{y}{x}$

Then $\displaystyle \frac{d}{dx}\Big(\frac{dy}{dx}\Big) = \frac{x\frac{dy}{dx}-y}{x^2}= 0$