# [SOLVED] Substitution help...

• February 6th 2010, 07:48 PM
snaes
[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:
$x^3 \dfrac{dy}{dx} + x^2y+x=3e^{xy}$

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

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

$\dfrac{dz}{dx}=ye^{xy}\dfrac{dy}{dx}$
• February 6th 2010, 08:12 PM
dedust
Quote:

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

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

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

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

if $z = e^{xy}$, then $\frac{dz}{dx} = e^{xy}\left(y + x\frac{dy}{dx}\right)$
• February 6th 2010, 08:15 PM
snaes
ahh
ahh with respect to both x and y.
THANKS!
• February 6th 2010, 09:17 PM
Prove It
Quote:

Originally Posted by snaes
ahh with respect to both x and y.
THANKS!

No, not with respect to both $x$ and $y$. Only with respect to $x$. But since $y$ is a function of $x$, you have to use the chain rule.

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

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

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

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

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

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