Question

I reallllllyyy need help, it's literally my last question and Im stuck on it.

f(0)=3
f'(0)=5
f"(0)=7
h(x)=cos(kx)[f(x)]+sinx
Find h'(x) at x=0.

Answer 1

We have

`h(x)=\cos (kx)*f(x)+\sin x`

We take the derivative of
`h(x)`:

`h'(x)=\cos (kx)*f'(x)+k(-\sin (kx))*f(x)+\cos x`

Now we simply plug in 0:

`h'(0)=\cos (k*0)*f'(0)-k\sin (k*0)*f(0)+\cos 0`

`=\cos 0 *5-k\sin 0 *3+\cos 0`

`=5*1-3k*0+1`

`=5+1`

`=6`