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Thread: [SOLVED] Substitution help...

  1. #1
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    [SOLVED] Substitution help...

    Hi, I just need help with taking the derivative of the substition factor. From there I can (hopefully) get the rest.

    DE:
    $\displaystyle x^3 \dfrac{dy}{dx} + x^2y+x=3e^{xy}$

    Change of variable given in problem:
    $\displaystyle z=e^{xy}$

    Is this part right, usually the right side of this only has a $\displaystyle y$, so im not quite sure what to do. Thanks.
    $\displaystyle \dfrac{d}{dx}[z=e^{xy}]$

    $\displaystyle \dfrac{dz}{dx}=ye^{xy}\dfrac{dy}{dx}$
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  2. #2
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    Quote Originally Posted by snaes View Post
    Hi, I just need help with taking the derivative of the substition factor. From there I can (hopefully) get the rest.

    DE:
    $\displaystyle x^3 \dfrac{dy}{dx} + x^2y+x=3e^{xy}$

    Change of variable given in problem:
    $\displaystyle z=e^{xy}$

    Is this part right, usually the right side of this only has a $\displaystyle y$, so im not quite sure what to do. Thanks.
    $\displaystyle \dfrac{d}{dx}[z=e^{xy}]$

    $\displaystyle \dfrac{dz}{dx}=ye^{xy}\dfrac{dy}{dx}$
    if $\displaystyle z = e^{xy}$, then $\displaystyle \frac{dz}{dx} = e^{xy}\left(y + x\frac{dy}{dx}\right)$
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  3. #3
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    ahh

    ahh with respect to both x and y.
    THANKS!
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  4. #4
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    Quote Originally Posted by snaes View Post
    ahh with respect to both x and y.
    THANKS!
    No, not with respect to both $\displaystyle x$ and $\displaystyle y$. Only with respect to $\displaystyle x$. But since $\displaystyle y$ is a function of $\displaystyle x$, you have to use the chain rule.


    I.e. $\displaystyle z = e^{xy}$

    Let $\displaystyle u = xy$ so that $\displaystyle z = e^u$.


    $\displaystyle \frac{du}{dx} = x\frac{d}{dx}(y) + y\frac{d}{dx}(x)$

    $\displaystyle = x\frac{dy}{dx} + y$.


    $\displaystyle \frac{dz}{du} = e^u = e^{xy}$.


    So $\displaystyle \frac{dz}{dx} = \left(x\frac{dy}{dx} + y\right)e^{xy}$.
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