Answer:
The answer is [tex]y^{'} =e^{\frac{1}{x} } (2x-1)[/tex]
Step-by-step explanation:
First of all, we have product of 2 functions [tex]e^{1/x}[/tex] and [tex]x^{2}[/tex]
[tex]y^{'} =u\frac{dv}{dx}+v\frac{du}{dx}[/tex]
Let [tex]u=e^{1/x}[/tex] and [tex]v=x^{2}[/tex]
using chain rule for [tex]u[/tex], [tex]\frac{du}{dx}=-\frac{1}{x^2}[/tex]
and [tex]\frac{dv}{dx}=2x[/tex]
Therefore,
[tex]y^{'} =u\frac{dv}{dx}+v\frac{du}{dx}=e^{1/x}(2x)+x^{2} (-\frac{1}{x^2})=e^{1/x}(2x-1)[/tex]
