Question

what is derivative of above image​

Answer 1

Answer:

e2y/x

Step-by-step explanation:

ey/x+y/x=e2y/x

Answer 2

Given,

$\longrightarrow\sf{y=e^{\frac{y}{x}}}$

Taking logarithm,

$\longrightarrow\sf{\log y=\log\left(e^{\frac{y}{x}}\right)}$

$\longrightarrow\sf{\log y=\dfrac{y}{x}\quad\quad\dots(1)}$

$\longrightarrow\sf{x=\dfrac{y}{\log y}}$

Now, differentiating with respect to $\sf{y,}$,

$\longrightarrow\sf{\dfrac{dx}{dy}=\dfrac{d}{dy}\left(\dfrac{y}{\log y}\right)}$

Applying quotient rule in the RHS,

$\longrightarrow\sf{\dfrac{dx}{dy}=\dfrac{1\cdot\log y-y\cdot\dfrac{1}{y}}{(\log y)^2}}$

$\longrightarrow\sf{\dfrac{dx}{dy}=\dfrac{\log y-1}{(\log y)^2}}$

From (1),

$\longrightarrow\sf{\dfrac{dx}{dy}=\dfrac{\dfrac{y}{x}-1}{\left(\dfrac{y}{x}\right)^2}}$

$\longrightarrow\sf{\dfrac{dx}{dy}=\dfrac{y-x}{x}\cdot\dfrac{x^2}{y^2}}$

$\longrightarrow\sf{\dfrac{dx}{dy}=\dfrac{x(y-x)}{y^2}}$

On taking the reciprocal we get,

$\longrightarrow\underline{\underline{\sf{\dfrac{dy}{dx}=\dfrac{y^2}{x(y-x)}}}}$