Question

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

Answer 1

`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.