Question

Let `y=[(x^(2x)(x-1)^3)/(3+5x)^4]`, how do you use logarithmic differentiation to find `dy/dx`?

Answer 1

`dy/dx= (x^(2x)(x-1)^3)/(3+5x)^4{2 + 2lnx + 3/(x-1)-20/(3+5x)}`

Explanation:

`y=[(x^(2x)(x-1)^3)/(3+5x)^4]`

Taking (natural) logs of both sides:

`ln y=ln[(x^(2x)(x-1)^3)/(3+5x)^4]`
`\ \ \ \ \ \=ln \ x^(2x) + ln\ (x-1)^3 - ln \ (3+5x)^4`
`\ \ \ \ \ \=2xln \ x + 3ln\ (x-1) - 4ln \ (3+5x)`

Differentiating the LHS implicitly and the RHS using the product rule gives:
`1/y dy/dx \ = (2x)(1/x) + (2)(lnx) + 3/(x-1)-4*5/(3+5x)`
`\ \ \ \ \ \ \ \ \ \ = 2 + 2lnx + 3/(x-1)-20/(3+5x)`
`:. dy/dx= (x^(2x)(x-1)^3)/(3+5x)^4{2 + 2lnx + 3/(x-1)-20/(3+5x)}`