evaluate the derivative of the following function.\nf(t) = (cos^-1 t)^5\nf(t) =

Explanation:

Step1: Apply the chain rule

Let (u = \cos^{-1}t), then (y = u^{5}). The chain rule states that (\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}). First, find (\frac{dy}{du}): If (y = u^{5}), then (\frac{dy}{du}=5u^{4}) (using the power rule (\frac{d}{du}(u^{n})=nu^{n - 1})).

Step2: Find (\frac{du}{dt})

If (u=\cos^{-1}t), then (\frac{du}{dt}=-\frac{1}{\sqrt{1 - t^{2}}}) (derivative formula for (y = \cos^{-1}x) is (y'=-\frac{1}{\sqrt{1 - x^{2}}})).

Step3: Substitute (u) back and multiply

Substitute (u = \cos^{-1}t) into (\frac{dy}{du}) and then multiply by (\frac{du}{dt}): (f'(t)=\frac{dy}{du}\cdot\frac{du}{dt}=5(\cos^{-1}t)^{4}\cdot\left(-\frac{1}{\sqrt{1 - t^{2}}}\right))

Answer:

(-\frac{5(\cos^{-1}t)^{4}}{\sqrt{1 - t^{2}}})