Question

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

Answer 1

`("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))`.