Question

How to differentiate y=xe^(-kx) using chain rule?
Please give step by step procedure

Answer 1

I like to think of chain rule as "derivative of the outside, times derivative of the inside" \[y=xe ^{-kx}\] but we see this is product rule as well, so derivative of x is 1, which gives us the first term will be \[e^{-kx}\] so we will end up with \[y'=e^{-kx}+x(e^{-kx})'\] whatever the derivative of e^-kx is... which is where we need chain rule. Remember, derivative of the outside times derivative of the inside.

The derivative of e^x is e^x.. but what's this?\[(-kx)'=?\] Another product rule!

Answer 2

\[y'=e^{-kx}+x(e^{-kx})\cdot(-kx)'\] Only one derivative left, I'll leave it to you.

Answer 3

Xeph,
I tend to think of the chain rule as:
(derivative of outside) • (inside) • (derivative of inside)
http://www.1728.org/chainrul.htm

Answer 4

The actual chain rule is\[f(g(x))=f'(g(x))\cdot g'(x)\] so I'd be careful we don't accidentally end up thinking that\[f(g(x))=f'(g(x))\cdot g(x)\cdot g'(x)\] which isn't correct (usually)

Answer 5

Although it's true (and I failed to mention before) that the chain rule is the derivative of the outside function AT THE INNER FUNCTION times the derivative of the inner function; thanks for catching that.

Answer 6

Thanks Xeph