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Chapter 3: Problem 9

Evaluate the derivatives of the following functions. $$f(w)=\cos \left(\sin ^{-1} 2 w\right)$$

Short Answer

Expert verified
Answer: The derivative of the function is $$f'(w) = \frac{-4w}{\sqrt{1 - 4w^2}}$$.

Step by step solution

01

Apply Chain Rule for cosine function

To start, consider \(f(w) = \cos(u)\), where \(u = \sin^{-1}(2w)\). Apply the chain rule: \(f'(w) = \frac{d}{dw}(\cos(u)) = -\sin(u) \frac{du}{dw}\) where we have made use of the fact that the derivative of \(\cos(u)\) is \(-\sin(u)\).
02

Apply Chain Rule for inverse sine function

Now, consider the derivative of \(u = \sin^{-1}(2w)\) with respect to \(w\). Apply the chain rule again: \(\frac{du}{dw} = \frac{1}{\sqrt{1 - (2w)^2}} \cdot 2 = \frac{2}{\sqrt{1 - 4w^2}}\) We used the mentioned derivative formula for the inverse sine function.
03

Substitute the derivative of \(u\) and Simplify

Substitute the derivative of \(u\) from step 2 into step 1: \(f'(w) = -\sin(u) \cdot \frac{2}{\sqrt{1 - 4w^2}}\) Recall that \(u = \sin^{-1}(2w)\). Therefore, \(-\sin(u) = -\sin(\sin^{-1}(2w)) = -2w\). Substitute this back into the expression for the derivative: \(f'(w) = -2w \cdot \frac{2}{\sqrt{1 - 4w^2}}\) Simplify the result: \(f'(w) = \frac{-4w}{\sqrt{1 - 4w^2}}\) So, the derivative of the given function is: $$f'(w) = \frac{-4w}{\sqrt{1 - 4w^2}}$$

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Most popular questions from this chapter

Calculate the following derivatives using the Product Rule. $$\begin{array}{lll} \text { a. } \frac{d}{d x}\left(\sin ^{2} x\right) & \text { b. } \frac{d}{d x}\left(\sin ^{3} x\right) & \text { c. } \frac{d}{d x}\left(\sin ^{4} x\right) \end{array}$$ d. Based upon your answers to parts (a)-(c), make a conjecture about \(\frac{d}{d x}\left(\sin ^{n} x\right),\) where \(n\) is a positive integer. Then prove the result by induction.

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