# Differenciate

• August 8th 2013, 05:56 AM
srirahulan
Differenciate
How can i differenciate $e^\frac{y}{x}=x$ respect to x>>>>>
• August 8th 2013, 06:09 AM
HallsofIvy
Re: Differenciate
Pretty straightforward, using the "chain rule". First, of course, we integrate functions, not equations, so what you really mean is "how do I differentiate each side of this equation?"

To differentiate $e^{\frac{y}{x}}$, let $u=\frac{y}{x}= yx^{-1}$. Then $\frac{de^u}{dx}= \frac{de^u}{du}\frac{du}{dx}$. $\frac{de^u}{du}= e^u$ and $\frac{dyx^{-1}}{dx}= (-1)yx^{-2}= -\frac{y}{x^2}$. So the derivative of $e^{\frac{y}{x}}$, with respect to x is $-\frac{y}{x^2}e^{\frac{y}{x}}$.

Now, what is the derivative of the right side of the equation?
• August 8th 2013, 06:33 AM
Soroban
Re: Differenciate
Hello, srirahulan!

Quote:

$\text{Differentiate: }\:e^\frac{y}{x}\:=\:x$

$\text{Take logs: }\:\ln\left(e^{\frac{y}{x}}\right) \:=\:\ln(x) \quad\Rightarrow\quad \frac{y}{x}\underbrace{\ln(e)}_{\text{This is 1}} \:=\:\ln(x)$

. . . . . . . . $\frac{y}{x} \:=\:\ln(x) \quad\Rightarrow\quad y \:=\:x\!\cdot\!\ln(x)$

$\text{Differentiate: }\:y' \;=\;x\!\cdot\!\frac{1}{x} + 1\!\cdot\!\ln(x)$

. . $\text{Therefore: }\:y' \;=\;1 + \ln(x)$
• August 8th 2013, 07:48 PM
ibdutt
Re: Differenciate