# How do you find the derivative e^(x+y) + e^2x +e^2y?

##### 1 Answer
May 23, 2015

I assume that you want to find $\frac{d}{\mathrm{dx}} \left({e}^{x + y} + {e}^{2 x} + {e}^{2 y}\right)$.

If I have misunderstood your question, please accept my apology.

$\frac{d}{\mathrm{dx}} \left({e}^{x + y} + {e}^{2 x} + {e}^{2 y}\right) = \frac{d}{\mathrm{dx}} \left({e}^{x + y}\right) + \frac{d}{\mathrm{dx}} \left({e}^{2 x}\right) + \frac{d}{\mathrm{dx}} \left({e}^{2 y}\right)$.

For each term, we will need the chain rule applied to the exponential function:

$\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dx}} \left({e}^{x + y} + {e}^{2 x} + {e}^{2 y}\right) = {e}^{x + y} \frac{d}{\mathrm{dx}} \left({e}^{x + y}\right) + {e}^{2 x} \frac{d}{\mathrm{dx}} \left(2 x\right) + {e}^{2 y} \frac{d}{\mathrm{dx}} \left(2 y\right)$.

$= {e}^{x + y} \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right) + 2 {e}^{2 x} + {e}^{2 y} \cdot 2 \frac{\mathrm{dy}}{\mathrm{dx}}$

$= {e}^{x + y} + {e}^{x + y} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 {e}^{2 x} + 2 {e}^{2 y} \frac{\mathrm{dy}}{\mathrm{dx}}$

So,
$\frac{d}{\mathrm{dx}} \left({e}^{x + y} + {e}^{2 x} + {e}^{2 y}\right) = {e}^{x + y} + {e}^{x + y} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 {e}^{2 x} + 2 {e}^{2 y} \frac{\mathrm{dy}}{\mathrm{dx}}$

If we had an equation, we could solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$, but as it is, all we can do is rewrite using algebra:

$\frac{d}{\mathrm{dx}} \left({e}^{x + y} + {e}^{2 x} + {e}^{2 y}\right) = \left[{e}^{x + y} + 2 {e}^{2 x}\right] + \left[{e}^{x + y} + 2 {e}^{2 y}\right] \frac{\mathrm{dy}}{\mathrm{dx}}$