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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