Question

How do you differentiate `f(x)=xe^(2x)-x` using the product rule?

Answer 1

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

Explanation:

using the `color(blue) " Product rule "`

If f(x) = g(x).h(x) then f'(x) = g(x).h'(x) + h(x).g'(x)

and `d/dx(e^x) = e^x`

the `color(blue)" chain rule " "is also used here "`

`d/dx[f(g(x))] = f'(g(x)). g'(x)`

hence : `f'(x) =[ x d/dx(e^(2x)) + e^(2x) d/dx(x)] - 1`

`=[ xe^(2x) d/dx(2x) + e^(2x).1 ] -1`

`=[ 2xe^(2x) + e^(2x) ] - 1 = e^(2x)(2x + 1) - 1`