Question

How do you use the quotien rule to differentiate `1/ (3x^2+2)^3`?

Answer 1

`dy/dx = (- 18x) / (3x^2+2)^4`

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.

We will need to differentiate the denominator (using the chain rule), so let's do that first :

`d/dx(3x^2+2)^3 = 3(3x^2+2)^2(6x)`
`:. d/dx(3x^2+2)^3 = 18x(3x^2+2)^2`

So with `y=1/(3x^2+2)^3` the quotient rule gives;

`dy/dx = ( ((3x^2+2)^3)(d/dx(1)) - (1)(d/dx(3x^2+2)^3) ) / ((3x^2+2)^3)^2`
`:. dy/dx = ( 0 - 18x(3x^2+2)^2 ) / (3x^2+2)^6`
`:. dy/dx = (- 18x) / (3x^2+2)^4`