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Thread: Implict differentiation

  1. #1
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    Implict differentiation

    Use implict diffeentiation to find dy/dx

    x^2(cosy)-y^2=e^2x
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  2. #2
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    Quote Originally Posted by Emmeyh15@hotmail.com View Post
    Use implict diffeentiation to find dy/dx

    x^2(cosy)-y^2=e^2x
    $\displaystyle \frac{d}{dx} ( x^2(\cos(y) - y^2) = \frac{d}{dx} e^{2x} $

    $\displaystyle \to \frac{d}{dx} ( x^2(\cos(y)) - \frac{d}{dx}(y^2) = \frac{d}{dx} e^{2x} $

    The first term requies the product rule!

    $\displaystyle \to x^2 \frac{d}{dx} (\cos(y)) + \cos(y)\frac{d}{dx}(x^2) - \frac{d}{dx}(y^2) = \frac{d}{dx} e^{2x} $

    The RHS can be written:

    $\displaystyle \to x^2 \frac{d}{dx} (\cos(y)) + \cos(y)\frac{d}{dx}(x^2) - \frac{d}{dx}(y^2) = \frac{d}{dx} (e^{x})^2 $

    And hence requires the chain rule:

    $\displaystyle \to x^2 \frac{d}{dx} (\cos(y)) + \cos(y)\frac{d}{dx}(x^2) - \frac{d}{dx}(y^2) = 2e^x \times \frac{d}{dx} (2x) $

    The first term also requires the chain rule:

    [tex] $\displaystyle \to x^2 (\frac{d}{dy} (\cos(y)) \times \frac{d}{dx}(y)) + \cos(y)\frac{d}{dx}(x^2) - \frac{d}{dx}(y^2) = 2e^{x} \times \frac{d}{dx} (2x) $

    Now all you have to do is carry out these relatively simple differentiations, and then change the equation so that you get $\displaystyle \frac{dy}{dx} = $.
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  3. #3
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    im confused on how to apply the chain rule to that? i got the poblem right so far but i'm stuck at that point
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  4. #4
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    Quote Originally Posted by Emmeyh15@hotmail.com View Post
    im confused on how to apply the chain rule to that? i got the poblem right so far but i'm stuck at that point
    well the first term on the LHS contains:

    $\displaystyle \frac{d}{dx} \cos(y) $

    The chain rule says that if you have a function $\displaystyle h(x) = f(g(x)) $, then $\displaystyle h'(x) = f'(g(x)) \times g'(x) $

    In our case $\displaystyle g(x) = y $, and $\displaystyle f(g(x)) = \cos(g(x)) $

    And hence $\displaystyle g'(x) = \frac{dy}{dx} $ and $\displaystyle f'(g(x)) = -\sin(g(x)) $

    Hence
    $\displaystyle \frac{d}{dx} \cos(y) = -\sin(y) \times \frac{dy}{dx} $

    In the RHS, we have the same sort of idea.

    $\displaystyle g(x) = e^x $ and $\displaystyle f(g(x)) = (g(x))^2 $

    Hence:

    $\displaystyle g'(x) = e^x $, and $\displaystyle f'(g(x)) = 2g(x) $

    So the result is :

    $\displaystyle frac{d}{dx} (e^x)^2 = e^x \times 2e^x = 2(e^x)^2 = 2e^{2x} $
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