Question

How do you find the derivative of `f(x) = [(3x^2 + 1)^(1/3) - 5]^2 / [5x^2 + 4]^(1/2)`?

Answer 1

`f'(x)=[(4x(5x^2+4)^(1/2))/((3x^2+1)^(2/3)){(3x^2+1)^(1/3)-5}-(5x)/(5x^2+4)^(1/2){(3x^2+1)^(1/3)-5}^2]/(5x^2+4)`

Explanation:

We will use the Quotient Rule and Chain Rule.

`f(x)={(3x^2+1)^(1/3) -5}^2/(5x^2+4)^(1/2)=u/v, say,`, where,

`u={(3x^2+1)^(1/3)-5}^2, &, v=(5x^2+4)^(1/2)`.

Now, `f=u/v rArr f'=(vu'-uv')/v^2...............(star)`.

`u={(3x^2+1)^(1/3)-5}^2`

`rArr u'=2{(3x^2+1)^(1/3)-5}{(3x^2+1)^(1/3)-5}'`

`=2{(3x^2+1)^(1/3)-5}{(3x^2+1)^(1/3)}'`

`=2{(3x^2+1)^(1/3)-5}{1/3(3x^2+1)^(-2/3)}(3x^2+1)'`

`=2/3*6x{(3x^2+1)^(1/3)-5}(3x^2+1)^(-2/3)`

`u'=(4x)/(3x^2+1)^(2/3){(3x^2+1)^(1/3)-5}`

`v=(5x^2+4)^(1/2)`.

`rArr v'=1/2(5x^2+4)^(-1/2)(5x^2+4)'`

`=1/2*10x(5x^2+4)^(-1/2)`

`v'=(5x)/(5x^2+4)^(1/2)`

Using all these, together with, `v^2=(5x^2+4)`, in `(star)`, we get,

`f'(x)=[(4x(5x^2+4)^(1/2))/((3x^2+1)^(2/3)){(3x^2+1)^(1/3)-5}-(5x)/(5x^2+4)^(1/2){(3x^2+1)^(1/3)-5}^2]/(5x^2+4)`

Enjoy maths.!