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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.

1 Answer

3 votes
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
User Imcoddy
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