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Thread: Need help with implicit differentiation

  1. #1
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    Need help with implicit differentiation

    Hi, I'm having an issue using implicit differentiation. I don't think I'm doing it correctly on the first step.

    Equation: "Find $\displaystyle dy/dx$ by implicit differentiation and evaluate the derivative at the given point"
    $\displaystyle tan (x+y) =x , (0,0)$

    What I have:
    $\displaystyle
    sec^2(x+y) (1+y') = (1)$
    $\displaystyle
    sec^2(x+y)+y'sec^2(x+y)
    $
    $\displaystyle
    y'=\frac {1-sec^2(x+y)}{sec^2(x+y)}$

    however, the answer for the first part is $\displaystyle -sin^2(x+y)$. Thanks in advance.
    Last edited by cynlix; Oct 7th 2009 at 04:54 PM.
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  2. #2
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    Quote Originally Posted by cynlix View Post
    Hi, I'm having an issue using implicit differentiation. I don't think I'm doing it correctly on the first step.

    Equation: "Find $\displaystyle dy/dx$ by implicit differentiation and evaluate the derivative at the given point"
    $\displaystyle tan (x+y) =x , (0,0)$

    What I have:
    $\displaystyle
    sec^2(x+y) (1+y') = (1)$
    $\displaystyle
    sec^2(x+y)+y'sec^2(x+y)
    $
    $\displaystyle
    y'=\frac {1-sec^2(x+y)}{sec^2(x+y)}$

    however, the answer for the first part is $\displaystyle -sin^2(x+y)$. Thanks in advance.
    same thing ...

    $\displaystyle \frac{1-\sec^2(x+y)}{\sec^2(x+y)}$

    multiply numerator and denominator by $\displaystyle \cos^2(x+y)$ ...

    $\displaystyle \frac{\cos^2(x+y) - 1}{1}$

    $\displaystyle -[1 - \cos^2(x+y)]$

    $\displaystyle -\sin^2(x+y)$

    ... review your trig identities.
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