1. ## Differenciate

How can i differenciate $\displaystyle e^\frac{y}{x}=x$ respect to x>>>>>

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

Now, what is the derivative of the right side of the equation?

3. ## Re: Differenciate

Hello, srirahulan!

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

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

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

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

. .$\displaystyle \text{Therefore: }\:y' \;=\;1 + \ln(x)$