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

Question

Elijah Abram · Accepted Answer

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

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))#.

Charles Anctil · Answer

undefined To find the derivative of the function $ y = (\cos^{-1}(4x^2))^2 $, we can use the chain rule and the power rule for differentiation.

Let $ u = \cos^{-1}(4x^2) $. Then $ y = u^2 $.

Using the chain rule, the derivative of $ u $ with respect to $ x $ is:
$$ \frac{du}{dx} = \frac{d}{dx}(\cos^{-1}(4x^2)) = -\frac{1}{\sqrt{1-(4x^2)^2}} \cdot \frac{d}{dx}(4x^2) $$
$$ = -\frac{1}{\sqrt{1-16x^4}} \cdot 8x $$

Now, using the power rule, the derivative of $ y $ with respect to $ u $ is:
$$ \frac{dy}{du} = \frac{d}{du}(u^2) = 2u $$

Finally, we combine these derivatives using the chain rule:
$$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2u \cdot \left(-\frac{1}{\sqrt{1-16x^4}} \cdot 8x\right) $$
$$ = -\frac{16xu}{\sqrt{1-16x^4}} $$

Substituting back $ u = \cos^{-1}(4x^2) $:
$$ = -\frac{16x\cos^{-1}(4x^2)}{\sqrt{1-16x^4}} $$