What is the first and second derivative of #(10x*(x-1)^4)/((x-2)^3*(x+1)^2) #?

1 Answer
Yiu A.
Oct 31, 2017

#dy/dx=[1/x+4/(x-1)-3/(x-2)-2/(x+1)] *[10x*(x-1)^4]/[(x-2)^3*(x+1)^2]#

Explanation:

As there are many power in this question and it is hard to differentiate this form, we can use the logarithmic differentiation to determine the derivative.

First, we know that

  1. #ln(a*b)= ln a+ ln b#
  2. #ln(a/b)= ln a- ln b#
  3. #ln(a^n)= nln a#

We can let #y=[10x*(x-1)^4]/[(x-2)^3*(x+1)^2]#
Then, #ln y=ln(10x)+ln[(x-1)^4]-{[ln(x-2)^3]+ln(x+1)^2}#

#ln y=ln(10x)+4ln(x-1)-[3ln(x-2)+2ln(x+1)]#

#ln y=ln(10x)+4ln(x-1)-3ln(x-2)-2ln(x+1)#

It is more easy for us to differentiate this form now.
And we also know that #d/dx(ln x)=1/x*d/dx(x)=1/x#

So, we can differentiate both sides to get the answer:
#d/dx(ln y)=d/dx[ln(10x)+4ln(x-1)-3ln(x-2)-2ln(x+1)]#

#1/y*dy/dx=1/(10x)*d/dx(10x)+4*1/(x-1)*d/dx(x-1)-3*1/(x-2)*d/dx(x-2)-2*1/(x+1)*d/dx(x+1)#

#1/y*dy/dx=1/(10x)*10+4/(x-1)*1-3/(x-2)*1-2/(x+1)*1#

#dy/dx=[1/x+4/(x-1)-3/(x-2)-2/(x+1)] *y#

#dy/dx=[1/x+4/(x-1)-3/(x-2)-2/(x+1)] *[10x*(x-1)^4]/[(x-2)^3*(x+1)^2]#

This is the answer for this question :)
Hope it can help you.

Related questions
See all questions in Quotient Rule
Impact of this question
1847 views around the world
You can reuse this answer
Creative Commons License