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Math Help - Derivative of Trig Function

  1. #1
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    Derivative of Trig Function

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

    Is this correct?
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  2. #2
    Member RedBarchetta's Avatar
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    No. Apply the product rule:

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

    Let u=(theta+1) and let v=cos(theta)
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  3. #3
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    Quote Originally Posted by RedBarchetta View Post
    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.
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  4. #4
    Member RedBarchetta's Avatar
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    Quote Originally Posted by elpermic View Post
    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.

    <br />
\begin{gathered}<br />
  f(\theta ) = (\theta  + 1)\cos \theta  \hfill \\<br />
  u = (\theta  + 1) \hfill \\<br />
  v = \cos \theta  \hfill \\<br />
  f(\theta ) = u \cdot v \hfill \\<br />
  f'(\theta ) = u'v + uv' \hfill \\<br />
  f'(\theta ) = 1 \cdot \cos \theta  + (\theta  + 1) \cdot  - \sin \theta  \hfill \\<br />
  f'(\theta ) = \cos \theta  - \theta \sin \theta  - \sin \theta  \hfill \\ <br />
\end{gathered} <br />
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  5. #5
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    Alright.. thanks!

    Got cos θ -sin^2 θ -1.
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