Question

Calculate y' using chain rule. y = (e ^ (1 / x))/(x ^ 2)

Answer 1

Answer:

The answer is $y^{'} =e^{\frac{1}{x} } (2x-1)$

Step-by-step explanation:

First of all, we have product of 2 functions $e^{1/x}$ and $x^{2}$

$y^{'} =u\frac{dv}{dx}+v\frac{du}{dx}$

Let $u=e^{1/x}$ and $v=x^{2}$

using chain rule for $u$, $\frac{du}{dx}=-\frac{1}{x^2}$

and $\frac{dv}{dx}=2x$

Therefore,

$y^{'} =u\frac{dv}{dx}+v\frac{du}{dx}=e^{1/x}(2x)+x^{2} (-\frac{1}{x^2})=e^{1/x}(2x-1)$