HELP plz!!!?

Question

Find the derivative of the function:

y=sin(3x^1/3)+(sin(4x))^1/3

Answer 1

y=sin(3x^1/3)+(sin(4x))^1/3

By chain rule

d/dx[ sin(3x^1/3) ]

= {cos(3x^1/3)}*d/dx{(3x^1/3)}

Now, d/dx{(3x^1/3)} = 3*(1/3)*x^(1/3 - 1) = x^(-2/3)

So, d/dx[ sin(3x^1/3) ] = x^(-2/3)cos(3x^1/3)



d/dx[ (sin(4x))^1/3 ]

= (1/3)*{sin(4x)}^(1/3 - 1)*d/dx[sin(4x)]

= (1/3)*{sin(4x)}^(-2/3)*4*cos(4x)

Hence, dy/dx = x^(-2/3)cos(3x^1/3) + (1/3)*{sin(4x)}^( -2/3)*4*cos(4x)