# What are the partial derivatives of cos(e^x+y) ?

Apr 26, 2017

If $f \left(x , y\right) = \sin \left({e}^{x} + y\right)$

${f}_{x} = {e}^{x} \cos \left({e}^{x} + y\right)$
${f}_{y} = \cos \left({e}^{x} + y\right)$

Whereas, if:

$f \left(x , y\right) = \sin \left({e}^{x + y}\right)$

Then:

${f}_{x} = {e}^{x + y} \cos \left({e}^{x + y}\right)$
${f}_{y} = {e}^{x + y} \cos \left({e}^{x + y}\right)$

#### Explanation:

I am non sure if your mean

$f \left(x , y\right) = \sin \left({e}^{x} + y\right)$, or
$f \left(x , y\right) = \sin \left({e}^{x + y}\right)$

So I will consider both. Remember when partially differentiating that we differentiate wrt the variable in question whilst treating the other variables as constant. And so:

If:

$f \left(x , y\right) = \sin \left({e}^{x} + y\right)$

Then:

${f}_{x} = \frac{\partial f}{\partial x}$
$\setminus \setminus \setminus = \frac{\partial}{\partial x} \left(\sin \left({e}^{x} + y\right)\right)$
$\setminus \setminus \setminus = \cos \left({e}^{x} + y\right) \frac{\partial}{\partial x} \left({e}^{x} + y\right)$
$\setminus \setminus \setminus = \cos \left({e}^{x} + y\right) \left({e}^{x}\right)$
$\setminus \setminus \setminus = {e}^{x} \cos \left({e}^{x} + y\right)$

${f}_{y} = \frac{\partial f}{\partial y}$
$\setminus \setminus \setminus = \frac{\partial}{\partial y} \left(\sin \left({e}^{x} + y\right)\right)$
$\setminus \setminus \setminus = \cos \left({e}^{x} + y\right) \frac{\partial}{\partial y} \left({e}^{x} + y\right)$
$\setminus \setminus \setminus = \cos \left({e}^{x} + y\right) \left(1\right)$
$\setminus \setminus \setminus = \cos \left({e}^{x} + y\right)$

Whereas, if:

$f \left(x , y\right) = \sin \left({e}^{x + y}\right)$

Then:

${f}_{x} = \frac{\partial f}{\partial x}$
$\setminus \setminus \setminus = \frac{\partial}{\partial x} \left(\sin \left({e}^{x + y}\right)\right)$
$\setminus \setminus \setminus = \cos \left({e}^{x + y}\right) \frac{\partial}{\partial x} \left({e}^{x + y}\right)$
$\setminus \setminus \setminus = \cos \left({e}^{x + y}\right) \left({e}^{x + y}\right)$
$\setminus \setminus \setminus = {e}^{x + y} \cos \left({e}^{x + y}\right)$

${f}_{y} = \frac{\partial f}{\partial y}$
$\setminus \setminus \setminus = \frac{\partial}{\partial y} \left(\sin \left({e}^{x + y}\right)\right)$
$\setminus \setminus \setminus = \cos \left({e}^{x + y}\right) \frac{\partial}{\partial y} \left({e}^{x + y}\right)$
$\setminus \setminus \setminus = \cos \left({e}^{x + y}\right) \left({e}^{x + y}\right)$
$\setminus \setminus \setminus = {e}^{x + y} \cos \left({e}^{x + y}\right)$