Question

How do you differentiate `root3(x)+root4(x)`?

Answer 1

`=1/(3x^(2/3))+1/(4x^(3/4))`.

Explanation:

We know that, `d/dx(x^n)=n*x^(n-1)`, and,

`d/dx(u+v)=(du)/dx+(dv)/dx`

Therefore, `d/dx(root3x+root4x)=d/dxroot3x+d/dxroot4x`

`=d/dxx^(1/3)+d/dxx^(1/4)`

`=1/3*x^(1/3-1)+1/4*x^(1/4-1)`

`=1/(3x^(2/3))+1/(4x^(3/4))`.