[SOLVED] Substitution help...

**snaes** · February 6th 2010, 06:48 PM

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

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

**dedust** · February 6th 2010, 07:12 PM

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

**snaes** · February 6th 2010, 07:15 PM

ahh with respect to both x and y.
THANKS!

**Prove It** · February 6th 2010, 08:17 PM

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

Thread: [SOLVED] Substitution help...

LinkBack

Thread Tools

Display

[SOLVED] Substitution help...

ahh

Similar Math Help Forum Discussions

[SOLVED] Help with trigonometric substitution

[SOLVED] Using both u and trig substitution

[SOLVED] Trig substitution

[SOLVED] Linear substitution

[SOLVED] substitution

Search Tags