Question

How do I solve this equation?

If `f(x)=(x-1)^2 sinx`, then `f'(0)=?`

Answer 1

`f'(0)=1`

Explanation:

we have to use the product rule

`f(x)=color(red)(u)v=>f'(x)=color(red)(u')v+color(red)(u)v'`

`f(x)=(x-1)^2sinx`

`color(red)(u=(x-1)^2=>u'=2(x-1))`

`v=sinx=>v'=cosx`

`:.f'(x)=color(red)(2(x-1))sinx+color(red)((x-1)^2)cosx`

`:. f'(0)=cancel(2(0-1)sin0)+(0-1)^2cos0`

`f'(0)=1xx1=1`