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Chapter 3: Problem 6

Explain how to find \(\left(f^{-1}\right)^{\prime}\left(y_{0}\right),\) given that \(y_{0}=f\left(x_{0}\right)\)

Short Answer

Expert verified
Answer: The general expression for the derivative of an inverse function \((f^{-1})'(y_0)\) given that \(y_0 = f(x_0)\) is \((f^{-1})'(y_0) = \frac{1}{f'(x_0)}\).

Step by step solution

01

Understand the inverse function and its notation

First, let's understand what an inverse function is. If \(f(x)\) is a function, its inverse function \(f^{-1}(y)\) is defined such that \(f(f^{-1}(y)) = y\). Notice that, in this case, the input variable of the inverse function is labeled with \(y\). Also, we know that if \(f(x_0) = y_0\), then \(f^{-1}(y_0) = x_0\).
02

Learn the formula for the derivative of an inverse function

The formula for computing the derivative of the inverse function \((f^{-1})'(y)\) is given by $$(f^{-1})'(y) = \frac{1}{f'(x)}$$ where \(x = f^{-1}(y)\).
03

Use the provided information to find the derivative

We are given that \(y_0 = f(x_0)\), which means that \(x_0 = f^{-1}(y_0)\). Therefore, we can substitute these values into the derivative formula: $$(f^{-1})'(y_0) = \frac{1}{f'(x_0)}$$
04

Conclusion

By using the provided value \(y_0 = f(x_0)\) along with the formula for the derivative of an inverse function, we were able to find a general expression for \((f^{-1})'(y_0)\). The final expression we derived was \((f^{-1})'(y_0) = \frac{1}{f'(x_0)}\).

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. \(\frac{d}{d x}\left(\sin ^{-1} x+\cos ^{-1} x\right)=0\) b. \(\frac{d}{d x}\left(\tan ^{-1} x\right)=\sec ^{2} x\) c. The lines tangent to the graph of \(y=\sin ^{-1} x\) on the interval [-1,1] have a minimum slope of 1 d. The lines tangent to the graph of \(y=\sin x\) on the interval \([-\pi / 2, \pi / 2]\) have a maximum slope of 1 e. If \(f(x)=1 / x,\) then \(\left[f^{-1}(x)\right]^{\prime}=-1 / x^{2}\)

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