1. ## [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.
$\displaystyle \dfrac{d}{dx}[z=e^{xy}]$

$\displaystyle \dfrac{dz}{dx}=ye^{xy}\dfrac{dy}{dx}$

2. Originally Posted by snaes
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 $\displaystyle \frac{dz}{dx} = e^{xy}\left(y + x\frac{dy}{dx}\right)$

3. ## ahh

ahh with respect to both x and y.
THANKS!

4. Originally Posted by snaes
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 $\displaystyle \frac{dz}{dx} = \left(x\frac{dy}{dx} + y\right)e^{xy}$.