Question

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

Answer 1

The answer is `=e^x(2-x^2)+(x)(x-2)sec^2x+2(x-1)tanx`

Explanation:

We need

`(e^x)'=e^x`

`(tanx)'=sec^2x`

The product rule is

`(uv)'=u'v+uv'`

Here,

`u=-e^x+tanx`, `=>`, `u'=-e^x+sec^2x`

`v=x^2-2x`, `=>`, `v'=2x-2`

Therefore,

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

`=-x^2e^x+2xe^x+x^2sec^2x-2xsec^2x-2xe^x+2e^x+2xtanx-2tanx`

`=e^x(2-x^2)+(x)(x-2)sec^2x+2(x-1)tanx`

Answer 2

`e^x(2-x^2)+2tanx(x-1)+x(x+2)sec^2x`

Explanation:

We know that,
`y=u*v,` then, `(dy)/(dx)=u*(dv)/(dx)+v*(du)/(dx)`
Taking, `u=-e^x+tanxand v=x^2-2x`we get
`(du)/(dx)=-e^x+sec^2xand(dv)/(dx)=2x-2`
`:.(dy)/(dx)=(-e^x+tanx)(2x-2)+(x^2-2x)(-e^x+sec^2x)`
`=cancel(-2xe^x)+2e^x+2xtanx-2tanx-x^2e^x+x^2sec^2x+cancel(2xe^x)+2xsec^2x`
`=e^x(2-x^2)+2(x-1)tanx+x(x+2)sec^2x`