f(θ)= (θ+1)cosθ

f'(θ)=(1)-sinθ

f"9θ)=-sinθ

Is this correct?

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- Sep 28th 2008, 04:15 PMelpermicDerivative of Trig Function
f(

**θ)= (****θ+1)cos****θ**

f'(**θ)=(1)-sin****θ**

f"9**θ)=-sin****θ**

Is this correct?

- Sep 28th 2008, 04:22 PMRedBarchetta
No. Apply the product rule:

f(x)=uv

f'(x)=u'v+uv'

Let u=(theta+1) and let v=cos(theta) - Sep 28th 2008, 04:25 PMelpermic
- Sep 28th 2008, 04:28 PMRedBarchetta
- Sep 28th 2008, 04:32 PMelpermic
Alright.. thanks!

Got cos**θ -sin^2****θ -1.**