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

1 Answer
1s2s2p
Jan 5, 2018

f'(x)=(pi(-e^(1/x)+5x(e^(1/x)+1)))/(x^7(e^(1/x)-1)^2)

Explanation:

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)

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