Question

Find derivative? `f(x) = (xe^x)^2`

Answer 1

`f^'(x)= 2 x e^(2 x)(x + 1)`

Explanation:

Differentiation of product of two functions:

`d/ (d x) (f(x)g(x)) = f^'(x) g(x) + f(x) g'(x)`

`f(x) = (x e^x)^2 or f(x) = x^2 * e^(2 x)`

`f^'(x)= x^2 * e^(2 x)* 2 + 2 x * e^(2 x)`

`:. f^'(x)= 2 x e^(2 x)(x + 1)` [Ans]

Answer 2

`f'(x)=2x(x+1)e^(2x)`

Explanation:

Here,

`f(x)=(xe^x)^2`

Let,

`y=u^2 ,where, color(red)(u=x*e^x)`

`:.(dy)/(du)=2u .......to(1)` .

`" Using "color(blue)"Product Rule"` , we get

`(du)/(dx)=x*d/(dx)(e^x)+e^x*d/(dx)(x)`

`=>(du)/(dx)=x*e^x+e^x*1...to(2)`

Now ,`"using "color(blue)"Chain Rule":`

`color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dx)`

`(dy)/(dx)=(2color(red)(u))xx(xe^x+e^x)....toFrom (1) and (2)`

`:.(dy)/(dx)=2color(red)((xe^x))xx(xe^x+e^x)`

`(dy)/(dx)=2xe^x xx e^x(x+1)`

`(dy)/(dx)=2x(x+1)e^(2x)`
...........................................................................................................

OR

`f(x)=(xe^x)^2=x^2e^(2x)`

`=>f'(x)=x^2d/(dx)(e^(2x))+e^(2x)d/(dx)(x^2)`

`=>f'(x)=x^2e^(2x)*2+e^(2x)(2x)=2x^2e^(2x)+2xe^(2x)`

Hence,

`f'(x)=(2x^2+2x)e^(2x)=2x(x+1)e^(2x)`