# Thread: find the derivative of a function

1. ## find the derivative of a function

find the derivative of a function

1. y= e^(-5x)cos(3x)

2. y=e^(x cos x)

????

find the derivative of a function

1. y= e^(-5x)cos(3x)

2. y=e^(x cos x)

????
You need to use product rule and chain rule for both:

1. $y=e^{-5x}\cos(3x)$

We see that $y'=-e^{-5x}\sin(3x)\cdot(3)+e^{-5x}\cdot(-5)\cdot\cos(3x)\implies \color{red}\boxed{y'=-3e^{-5x}\sin(3x)-5e^{-5x}\cos(3x)}$

I leave #2 for you to do.

Note that $\frac{d}{\,dx}\left[e^u\right]=e^u\cdot\frac{\,du}{\,dx}$. This means that you will end up applying the product rule to $x\cos x$.

Can you take this one from here?

--Chris

find the derivative of a function

1. y= e^(-5x)cos(3x)

2. y=e^(x cos x)

????
1. $y=e^{-5x}\cos{3x}$

You have a product of functions, so use the product rule.

$\frac{dy}{dx}=e^{-5x}\frac{d}{dx}(\cos{3x}) + \frac{d}{dx}\left(e^{-5x}\right)\cos{3x}$

$\frac{dy}{dx}=-3e^{-5x}\sin{3x} - 5e^{5x}\cos{3x}$

2. $y = e^{x \cos{x}}$.

This is a composition of functions, so use the chain rule.

Let $u = x \cos{x}$, so $y=e^u$.

$\frac{du}{dx} = -x\sin{x} + \cos{x}$ (use the product rule).

$\frac{dy}{du} = e^u = e^{x \cos{x}}$.

$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$

$= e^{x \cos{x}}(-x \sin{x} + \cos{x})$.