How do you differentiate #f(x) =x^2/(e^(x)-1)# using the quotient rule?

Question

Cameron Alexander · Accepted Answer

#=( x((2- x) e^x -2))/(e^(x)-1)^2# Rule of Quotient #d(u/v) = (d(u) v - u d(v))/v^2# here #d (x^2/(e^(x)-1))# #=(2x(e^(x)-1) - x^2 e^x)/(e^(x)-1)^2# #=( x((2- x) e^x -2))/(e^(x)-1)^2#

Ezra Adams · Answer

undefined To differentiate the function $f(x) = \frac{x^2}{e^x - 1}$ using the quotient rule, we'll follow the quotient rule formula which is $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $u = x^2$ and $v = e^x - 1$.

First, we need to find the derivatives of $u$ and $v$ with respect to $x$.

1. For $u = x^2$, the derivative is $u' = 2x$.
2. For $v = e^x - 1$, the derivative is $v' = e^x$, since the derivative of $e^x$ is $e^x$ and the derivative of a constant is 0.

Substituting these into the quotient rule formula gives us:

$$
f'(x) = \frac{(2x)(e^x - 1) - (x^2)(e^x)}{(e^x - 1)^2}
$$

Expanding and simplifying inside the numerator:

$$
f'(x) = \frac{2xe^x - 2x - x^2e^x}{(e^x - 1)^2}
$$

$$
f'(x) = \frac{x(2e^x - 2 - xe^x)}{(e^x - 1)^2}
$$

This is the derivative of $f(x) = \frac{x^2}{e^x - 1}$ using the quotient rule.