What is the derivative of this function #y=(cos^-1(4x^2))^2#?

1 Answer
Aug 4, 2017

#("d")/("d"x) ("arccos"(4x^2)) = -(16x*"arccos"(4x^2))/(sqrt(1-16x^4))#.

Explanation:

Note that #"d"/("d"x) ("arccos"(x))=-1/sqrt(1-x^2)#.

Then, by the chain rule
#("d")/("d"x) ("arccos"(4x^2)) = 2*("arccos"(4x^2))*"d"/("d"x) ("arccos"(4x^2))#.

Then, by the chain rule again,
#"d"/("d"x) ("arccos"(4x^2))=-1/sqrt(1-16x^4) * "d"/("d"x) (4x^2)#,
#"d"/("d"x) ("arccos"(4x^2))=-(8x)/sqrt(1-16x^4)#.

Substituting,

#("d")/("d"x) ("arccos"(4x^2)) = -(16x*"arccos"(4x^2))/(sqrt(1-16x^4))#.