# Thread: Derivative of Trig Function

1. ## Derivative of Trig Function

f( θ)= ( θ+1)cos θ
f'(
θ)=(1)-sin θ
f"9
θ)=-sin θ

Is this correct?

2. No. Apply the product rule:

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

Let u=(theta+1) and let v=cos(theta)

3. Originally Posted by RedBarchetta
No. Apply the chain rule:

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

Let u=(theta+1) and let v=cos(theta)
What is the chain rule?? I havent learned of this yet.. SO far in my class we are doing Product and Quotient Rules.

4. Originally Posted by elpermic
What is the chain rule?? I havent learned of this yet.. SO far in my class we are doing Product and Quotient Rules.
Oops. Sorry about that! I meant use the 'product rule'. Which is exactly what I posted.

$\displaystyle \begin{gathered} f(\theta ) = (\theta + 1)\cos \theta \hfill \\ u = (\theta + 1) \hfill \\ v = \cos \theta \hfill \\ f(\theta ) = u \cdot v \hfill \\ f'(\theta ) = u'v + uv' \hfill \\ f'(\theta ) = 1 \cdot \cos \theta + (\theta + 1) \cdot - \sin \theta \hfill \\ f'(\theta ) = \cos \theta - \theta \sin \theta - \sin \theta \hfill \\ \end{gathered}$

5. Alright.. thanks!

Got cos θ -sin^2 θ -1.