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

1 Answer
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.