Question

How do you find the derivative of `f(x)=2/(3x^3-x^2-24x-4)`?

Answer 1

`f'(x) = -(2(9x^2-2x-24) ) / (3x^3-x^2-24x-4)^2`

Explanation:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

`d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2`, or less formally, `(u/v)' = (v(du)-u(dv))/v^2`

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with `f(x )= 2/(3x^3-x^2-24x-4)` Then

`{ ("Let "u=2, => , (du)/dx=0), ("And "v=3x^3-x^2-24x-4, =>, (dv)/dx=9x^2-2x-24 ) :}`

`:. d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2`
`:. f'(x) = ( (3x^3-x^2-24x-4)(0) - (2)(9x^2-2x-24) ) / (3x^3-x^2-24x-4)^2`
`:. f'(x) = -(2(9x^2-2x-24) ) / (3x^3-x^2-24x-4)^2`

Incidental, we could also use the chain rule as follows;

`f(x )= 2/(3x^3-x^2-24x-4)`
`:. f(x )= 2(3x^3-x^2-24x-4)^-1`
`:. f'(x )= 2(-1)(3x^3-x^2-24x-4)^-2 * d/dx(3x^3-x^2-24x-4)`
`:. f'(x )= -2/(3x^3-x^2-24x-4)^2 * d/dx(3x^3-x^2-24x-4)`
`:. f'(x )= -2/(3x^3-x^2-24x-4)^2 * (9x^2-2x-24)`
`:. f'(x )= -(2(9x^2-2x-24))/(3x^3-x^2-24x-4)^2`