# Math Help - differentiation of e using chain rule

1. ## differentiation of e using chain rule

Could someone assist with this, please? The task is to differentiate the following equation

$y=xe^{-kx}$

by the chain rule, differentiation should be $e^{f(x)}*f'(x)$ for e, so to me that means

$y'=xe^{-kx}(-k)$ or $y'=-kxe^{-kx}$

unfortunately, my book would disagree with me. According to it, the answer before simplification is

$y'=xe^{-kx}(-k) + e^{-kx}*1$

Can someone please explain to me where the plus and everything after it came from?

2. You need to use the product rule on the expression as a whole

If $y = u(x)v(x)$ then $y' = u'v + v'u$

You should set $u=x$ and $v = e^{-kx}$

Use the chain rule to find v'

3. Originally Posted by satis
Could someone assist with this, please? The task is to differentiate the following equation

$y=xe^{-kx}$

by the chain rule, differentiation should be $e^{f(x)}*f'(x)$ for e, so to me that means

$y'=xe^{-kx}(-k)$ or $y'=-kxe^{-kx}$

unfortunately, my book would disagree with me. According to it, the answer before simplification is

$y'=xe^{-kx}(-k) + e^{-kx}*1$

Can someone please explain to me where the plus and everything after it came from?
1. The given function is a product of functions. That means you have to use the product rule:

$f(x)=u(x) \cdot v(x)~\implies~ f'(x)=v(x) \cdot u'(x) + u(x) \cdot v'(x)$

2. With your function you have $u(x) = x, v(x) = e^{-kx}$

3. The first step should be:

$f'(x)=e^{-kx} \cdot 1 + x \cdot \underbrace{e^{-kx} \cdot (-k)}_{chain-rule}$

4. Simplify!

4. ahhh, got it. I'd totally glossed over the fact I had a product of two functions. thanks to both of you for pointing out the problem.

5. Originally Posted by satis
Could someone assist with this, please? The task is to differentiate the following equation

$y=xe^{-kx}$

by the chain rule, differentiation should be $e^{f(x)}*f'(x)$ for e, so to me that means

$y'=xe^{-kx}(-k)$ or $y'=-kxe^{-kx}$

unfortunately, my book would disagree with me. According to it, the answer before simplification is

$y'=xe^{-kx}(-k) + e^{-kx}*1$

Can someone please explain to me where the plus and everything after it came from?

$y' = (x)' e^{-kx} + x (e^{-kx} )'$

Edit: sorry ... to late