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

1 Answer
Mar 27, 2018

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))} #