To find the derivative of ( \arccos^2(4x) ), we can use the chain rule. Let's break it down step by step:
Let ( u = 4x )( \Rightarrow \arccos^2(4x) = \arccos^2(u) )
Now, we can differentiate with respect to ( u ) and then multiply by the derivative of ( u ) with respect to ( x ):
( y = \arccos^2(u) )
Apply chain rule to find ( dy/du ):
( dy/du = 2\arccos(u) \times \frac{1}{\sqrt{1-u^2}} )
Now, differentiate ( u ) with respect to ( x ):
( du/dx = 4 )
Finally, apply chain rule to get the derivative of the entire expression with respect to ( x ):
( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = 2\arccos(4x) \times \frac{1}{\sqrt{1-(4x)^2}} \times 4 )
So, the derivative of ( \arccos^2(4x) ) with respect to ( x ) is ( \frac{8\arccos(4x)}{\sqrt{1-16x^2}} ).
To find the derivative of ( \arccos^2(4x) ), we can use the chain rule. Let's break it down step by step:
Let ( u = 4x )
( \Rightarrow \arccos^2(4x) = \arccos^2(u) )
Now, we can differentiate with respect to ( u ) and then multiply by the derivative of ( u ) with respect to ( x ):
( y = \arccos^2(u) )
Apply chain rule to find ( dy/du ):
( dy/du = 2\arccos(u) \times \frac{1}{\sqrt{1-u^2}} )
Now, differentiate ( u ) with respect to ( x ):
( du/dx = 4 )
Finally, apply chain rule to get the derivative of the entire expression with respect to ( x ):
( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = 2\arccos(4x) \times \frac{1}{\sqrt{1-(4x)^2}} \times 4 )
So, the derivative of ( \arccos^2(4x) ) with respect to ( x ) is ( \frac{8\arccos(4x)}{\sqrt{1-16x^2}} ).