# How do you find the derivative of y=e^x cos(x) ?

Aug 5, 2014

This is a type of problem involving the product rule.

The product rule states:

$\frac{d}{\mathrm{dx}} \left[f \left(x\right) \cdot g \left(x\right)\right] = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

So, we will let $f \left(x\right) = {e}^{x}$, and $g \left(x\right) = \cos x$.

We know that the derivative of ${e}^{x}$ is simply ${e}^{x}$, and that the derivative of $\cos x$ is equal to $- \sin x$.

(if these identities look unfamiliar to you, I may recommend viewing videos from this page or this page, which explain the derivative rules for ${e}^{x}$ and $\cos x$ more in-depth)

Therefore, $f ' \left(x\right) = {e}^{x}$, and $g ' \left(x\right) = - \sin x$. We can then simply substitute into the product rule formula:

$\frac{d}{\mathrm{dx}} \left[{e}^{x} \cos x\right] = {e}^{x} \cdot \left(- \sin x\right) + {e}^{x} \cdot \cos x$

To make this equation a little prettier, we will factor the ${e}^{x}$:

$\frac{d}{\mathrm{dx}} \left[{e}^{x} \cos x\right] = {e}^{x} \cdot \left(\cos x - \sin x\right)$

Jun 14, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} \left(- \sin x\right) + \cos x \left({e}^{x}\right)$

#### Explanation:

When two variables are multiplied in derivative
We Have Formula,
$\frac{d \left(u v\right)}{\mathrm{dx}} = u . \left(\frac{d \left(v\right)}{\mathrm{dx}}\right) + v . \left(\frac{d \left(u\right)}{\mathrm{dx}}\right)$
Question: To find the derivative of $y = {e}^{x} \cos \left(x\right)$ ?

Diffentiating both side by $x$
We get :
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d \left({e}^{x} \cos \left(x\right)\right)}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} \left(\frac{d \left(\cos x\right)}{\mathrm{dx}}\right) + \cos x \left(\frac{d \left({e}^{x}\right)}{\mathrm{dx}}\right)$

Here,
$\frac{d \left(\cos x\right)}{\mathrm{dx}} = - \sin x$ for first derivative.

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} \left(- \sin x\right) + \cos x \left({e}^{x}\right)$

Which is required solution.