What is the derivative of #f(x)=(pi/x^5)(1/(e^(1/x)-1))#?

Question

Evelyn Agard · Accepted Answer

#f'(x)=(pi(-e^(1/x)+5x(e^(1/x)+1)))/(x^7(e^(1/x)-1)^2)# The equation can be simplified as #f(x)=pi/(x^5(e^(1/x)-1))=pi/g(x)# Using the quotient rule: #f(x)=(g(x))/(h(x))=>f'(x)=(h(x)g'(x)-h'(x)g(x))/(h(x))^2# We can get: #f'(x)=(-pig'(x))/(g(x))^2# since #d/(dx)[pi]=0# #g(x)=x^5(e^(1/x)-1)=h(x)j(x)# #g'(x)=h(x)j'(x)+h'(x)j(x)# #h(x)=x^5# #h'(x)=5x^4# #j(x)=e^(1/x)-1=e^(a(x))-1# #j'(x)=a'(x)e^(a(x))# #a(x)=x^(-1)# #a'(x)=-x^(-2)=-1/x^2# #j'(x)=-e^(1/x)/x^2# #g'(x)=x^5(-e^(1/x)/x^2)+5x^4(e^(1/x)-1)=-x^3e^(1/x)+5x^4(e^(1/x)-1)=x^3(-e^(1/x)+5x(e^(1/x)+1))# #f'(x)=(pix^3(-e^(1/x)+5x(e^(1/x)+1)))/(x^5(e^(1/x)-1))^2# #color(white)(Xllll)=(pix^3(-e^(1/x)+5x(e^(1/x)+1)))/(x^10(e^(1/x)-1)^2)# #color(white)(Xllll)=(pi(-e^(1/x)+5x(e^(1/x)+1)))/(x^7(e^(1/x)-1)^2)#

Luke Alvarez · Answer

undefined To find the derivative of $ f(x) = \frac{\pi}{x^5}\left(\frac{1}{e^{1/x}-1}\right) $, we will use the product rule and chain rule.

Let $ u(x) = \frac{\pi}{x^5} $ and $ v(x) = \frac{1}{e^{1/x}-1} $.

Using the product rule: $ f'(x) = u'(x)v(x) + u(x)v'(x) $.

Now, let's find the derivatives of $ u(x) $ and $ v(x) $:

$ u'(x) = \frac{d}{dx}\left(\frac{\pi}{x^5}\right) = -\frac{5\pi}{x^6} $.

$ v'(x) = \frac{d}{dx}\left(\frac{1}{e^{1/x}-1}\right) $.

To find $ v'(x) $, we'll use the chain rule. Let $ g(x) = e^{1/x} $.

$ v'(x) = \frac{d}{dx}\left(\frac{1}{g(x)-1}\right) = \frac{-1}{(g(x)-1)^2} \cdot g'(x) $.

Now, find $ g'(x) $:

$ g'(x) = \frac{d}{dx}(e^{1/x}) = e^{1/x} \cdot \frac{d}{dx}(1/x) = -\frac{1}{x^2} \cdot e^{1/x} $.

Substitute $ g(x) $ and $ g'(x) $ back into $ v'(x) $:

$ v'(x) = \frac{-1}{(e^{1/x}-1)^2} \cdot \left(-\frac{1}{x^2} \cdot e^{1/x}\right) $.

Now, put all the pieces together:

$ f'(x) = \frac{-5\pi}{x^6} \cdot \frac{1}{e^{1/x}-1} + \frac{\pi}{x^5} \cdot \frac{-1}{(e^{1/x}-1)^2} \cdot \left(-\frac{1}{x^2} \cdot e^{1/x}\right) $.

Simplify the expression to get the final derivative of $ f(x) $.