# Derivative of Trig Function

• Sep 28th 2008, 03:15 PM
elpermic
Derivative 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, 03:22 PM
RedBarchetta
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, 03:25 PM
elpermic
Quote:

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.
• Sep 28th 2008, 03:28 PM
RedBarchetta
Quote:

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.

$
\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}
$
• Sep 28th 2008, 03:32 PM
elpermic
Alright.. thanks!

Got cos θ -sin^2 θ -1.