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

Question

Aurora Alvarado · Accepted Answer

See solution below.

#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]# When we combine all of this, we get #d/{du}[u^{1/2}] times (d/dx[3x]-(d/{dv}[v^2] (d/{dw}[sinw]d/dx[4x])))# Assessing this expression, we obtain: #1/2 u^{-1/2} times (3-(2v (cosw times 4)))# Changing every value we receive: #1/2 (3x-(sin4x)^2)^{-1/2} times (3-(2(sin4x) (cos(4x) times 4)))# This is as simple as: # (3-8sin4xcos4x)/{2 times sqrt{(3x-sin^2(4x))} #

Cameron Alexander · Answer

undefined To find the derivative of the given function, we can use the chain rule and the power rule.

First, let's denote the given function as $ y = (3x - (\sin(4x))^2)^{\frac{1}{2}} $.

Now, we'll differentiate step by step:

1. Differentiate the outer function with respect to the inner function:

$$ \frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}} $$

2. Differentiate the inner function:

$$ \frac{du}{dx} = 3 - 2\sin(4x)\cdot4\cos(4x) $$

3. Apply the chain rule:

$$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $$

4. Substitute $ u $ with the inner function:

$$ \frac{dy}{dx} = \frac{1}{2}(3x - (\sin(4x))^2)^{-\frac{1}{2}} \cdot (3 - 2\sin(4x)\cdot4\cos(4x)) $$

This expression represents the derivative of the given function.