f(θ)= (θ+1)cosθ

f'(θ)=(1)-sinθ

f"9θ)=-sinθ

Is this correct?

Printable View

- Sep 28th 2008, 03:15 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, 03:22 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, 03:25 PMelpermic
- Sep 28th 2008, 03:28 PMRedBarchetta
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}

$ - Sep 28th 2008, 03:32 PMelpermic
Alright.. thanks!

Got cos**θ -sin^2****θ -1.**