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Chapter 3: Problem 11

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

Short Answer

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

Step by step solution

01

Find the derivative of the outer function

To find the derivative of $$g(x) = \sin^{-1}(x)$$, we can use the following formula: $$g'(x) = \frac{1}{\sqrt{1 - x^2}}$$ So the derivative of the outer function is: $$g'(h(x)) = \frac{1}{\sqrt{1 - h(x)^2}}$$
02

Find the derivative of the inner function

To find the derivative of $$h(x) = e^{-2x}$$, apply the chain rule for exponential functions: $$h'(x) = -2e^{-2x}$$
03

Apply the chain rule to find the derivative of the composite function

Now that we have $$g'(h(x))$$ and $$h'(x)$$, we can apply the chain rule to find the derivative of the composite function, $$f(x) = g(h(x))$$. $$f'(x) = g'(h(x)) \cdot h'(x)$$ So, substituting the values we found for $$g'(h(x))$$ and $$h'(x)$$: $$f'(x) = \frac{1}{\sqrt{1 - h(x)^2}} \cdot (-2e^{-2x})$$ $$f'(x) = \frac{1}{\sqrt{1 - (e^{-2x})^2}} \cdot (-2e^{-2x})$$ This gives us the final expression for the derivative of $$f(x)$$: $$f'(x) = \frac{-2e^{-2x}}{\sqrt{1 - e^{-4x}}}$$

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Most popular questions from this chapter

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