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Thread: implicit differentiation

  1. #1
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    implicit differentiation

    Given:
    $\displaystyle ye^y=(x-5y)^3$

    Find $\displaystyle \frac{dy}{dx}$, in terms of x and y.

    what I tried to do was

    Differentiate implicitly wrt x,
    $\displaystyle \frac{dy}{dx}(e^y)+\frac{dy}{dx}(y)(e^y)=(3)(x-5y)^2(1-5\frac{dy}{dx})$
    $\displaystyle \frac{dy}{dx}(e^y)(1+y)=3(x-5y)^2(1-5\frac{dy}{dx})$

    then, i'm stuck in moving all $\displaystyle \frac{dy}{dx}$ to the left-hand side.
    any help is much appreciated. thanks.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: implicit differentiation

    Quote Originally Posted by wintersoltice View Post
    Given:
    $\displaystyle ye^y=(x-5y)^3$

    Find $\displaystyle \frac{dy}{dx}$, in terms of x and y.

    what I tried to do was

    Differentiate implicitly wrt x,
    $\displaystyle \frac{dy}{dx}(e^y)+\frac{dy}{dx}(y)(e^y)=(3)(x-5y)^2(1-5\frac{dy}{dx})$
    $\displaystyle \frac{dy}{dx}(e^y)(1+y)=3(x-5y)^2(1-5\frac{dy}{dx})$

    then, i'm stuck in moving all $\displaystyle \frac{dy}{dx}$ to the left-hand side.
    any help is much appreciated. thanks.
    $\displaystyle \frac{dy}{dx}(e^y)(1+y)=3(x-5y)^2-15\frac{dy}{dx}(x-5y)^2$

    $\displaystyle \frac{dy}{dx}[e^y(1+y)+15(x-5y)^2]=3(x-5y)^2$

    $\displaystyle \frac{dy}{dx}=\frac{3(x-5y)^2}{e^y(1+y)+15(x-5y)^2}$
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  3. #3
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    Re: implicit differentiation

    Quote Originally Posted by alexmahone View Post
    $\displaystyle \frac{dy}{dx}(e^y)(1+y)=3(x-5y)^2-15\frac{dy}{dx}(x-5y)^2$

    $\displaystyle \frac{dy}{dx}[e^y(1+y)+15(x-5y)^2]=3(x-5y)^2$

    $\displaystyle \frac{dy}{dx}=\frac{3(x-5y)^2}{e^y(1+y)+15(x-5y)^2}$
    how did $\displaystyle (1-5\frac{dy}{dx})$
    become $\displaystyle -15\frac{dy}{dx}(x-5y)^2$?
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Re: implicit differentiation

    Quote Originally Posted by wintersoltice View Post
    how did $\displaystyle (1-5\frac{dy}{dx})$
    become $\displaystyle -15\frac{dy}{dx}(x-5y)^2$?
    $\displaystyle 3(1-5\frac{dy}{dx})=3-15\frac{dy}{dx}$
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