# Thread: How do you differentiate e^xy?

1. ## How do you differentiate e^xy?

Topic.

I took a stab at it, and my guess is that you use the chain rule, but it looks messed up in the end.
Any help would be greatly appreciated.

2. Originally Posted by cjh824
Topic.

I took a stab at it, and my guess is that you use the chain rule, but it looks messed up in the end.
Any help would be greatly appreciated.
Use the chain rule

$u = xy$ the n
$\frac{d}{dx} e^u = \frac{d}{du} e^u \cdot \frac{du}{dx} = e^u (x y'+y) = e^{xy}(xy'+y)$

3. Originally Posted by cjh824
Topic.

I took a stab at it, and my guess is that you use the chain rule, but it looks messed up in the end.
Any help would be greatly appreciated.
I assume you mean $e^{xy}$ rather than $ye^x$

Let $f(x) = e^{xy}$ and then taking logs of both sides (in base e) gives:

$\ln{f(x)} = xy$

Now we differentiate implicitly and with the product rule recalling that the differentiating $\ln{f(x)} = \frac{f\prime(x)}{f(x)}$:

Right hand side: u=x and v=y so du/dx = 1 and dv/dx = dy/dx which gives us the right hand side differential as $y + x\frac{dy}{dx}$

$\frac{f\prime(x)}{f(x)} = y + x\frac{dy}{dx}$

As we know f(x) = e^(xy) we can say that $\frac{f\prime(x)} = ye^{xy} + {xe^{xy}\frac{dy}{dx}$

We can now collect dy/dx terms (and in this case $f\prime{x}$ can also be treated as dy/dx)

$\frac{dy}{dx} - {xe^{xy}}\frac{dy}{dx} = ye^(xy)$

Factor on the left:

$\frac{dy}{dx}(1 - {xe^{xy}}) = ye^(xy)$

and finally divide by $(1 - {xe^{xy}})$

$\frac{dy}{dx} = \frac{ye^(xy)}{(1 - {xe^{xy}})}$

4. $\left(e^{xy}\right)'=e^{xy}(x'y+y'x)=e^{xy}(y+xy')$

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