Question

How do you find the derivative of `(3x-(sin4x)^2)^(1/2)`?

Answer 1

See solution below.

Explanation:

`d/dx[(3x-(sin4x)^2)^{1/2}]`

This one is a bit of a hassle but with proper substitution it's possible. Think of this expression as multiple functions.
`(3x-(sin4x)^2)^{1/2}` is a function of `f` where `f(x)=x^{1/2}` and `x=3x-(sin4x)^2`.
Always remember to multiply the derivative of the inner function by the derivative of the outer function. It's a bit hard for me to explain so please follow the process below.

`d/dx[(3x-(sin4x)^2)^{1/2}]`

Start by using the substitution: `u=3x-(sin4x)^2`

`d/dx[(3x-(sin4x)^2)^{1/2}]=d/{du}[u^{1/2}] times d/dx[3x-(sin4x)^2]`

`d/dx[3x-(sin4x)^2]= d/dx[3x]-d/dx[(sin4x)^2]`

`v=sin4x`

`d/dx[(sin4x)^2] = d/{dv}[v^2]d/dx[sin4x]`

`w=4x`

`d/dx[sin4x]=d/{dw}[sinw]d/dx[4x]`

Now putting all this together we get

`d/{du}[u^{1/2}] times (d/dx[3x]-(d/{dv}[v^2] (d/{dw}[sinw]d/dx[4x])))`

evaluating this expression we get:

`1/2 u^{-1/2} times (3-(2v (cosw times 4)))`

Substituting all the values we get:

`1/2 (3x-(sin4x)^2)^{-1/2} times (3-(2(sin4x) (cos(4x) times 4)))`

This can be simplified to:

`(3-8sin4xcos4x)/{2 times sqrt{(3x-sin^2(4x))}`