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Chapter 3: Problem 7

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

Short Answer

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

Step by step solution

01

Identify the Inner and Outer Functions

The given function \(f(x) = \sin^{-1}(2x)\) can be written as a composite function, with an inner function \(u(x) = 2x\) and an outer function \(y(u) = \sin^{-1}(u)\). This will help us in using the chain rule to solve for the derivative of \(f(x)\).
02

Differentiate the Inner Function

First, we need to differentiate the inner function, \(u(x) = 2x\). Using the power rule, we have: $$u'(x) = \frac{d(2x)}{dx} = 2$$
03

Differentiate the Outer Function

Next, we need to differentiate the outer function, \(y(u) = \sin^{-1}(u)\). Recall that the derivative of the inverse sine function is: $$\frac{dy(u)}{du} = \frac{1}{\sqrt{1-u^2}}$$ In this case, \(u=2x\). We will substitute \(u=2x\) into the derivative expression to get the derivative of the outer function with respect to \(x\): $$\frac{dy(2x)}{dx} = \frac{1}{\sqrt{1-(2x)^2}}$$
04

Apply the Chain Rule

Now we will apply the chain rule, which states that when we have a composite function, the derivative is given as follows: $$f'(x) = \frac{dy(u)}{du} \cdot \frac{du(x)}{dx}$$ Substituting the derivatives of the inner and outer functions, we get: $$f'(x) = \frac{1}{\sqrt{1-(2x)^2}} \cdot 2$$
05

Simplify the Derivative Expression

Finally, we can simplify the derivative expression: $$f'(x) = \frac{2}{\sqrt{1-4x^2}}$$ So, the derivative of the given function \(f(x) = \sin^{-1}(2x)\) is: $$f'(x) = \frac{2}{\sqrt{1-4x^2}}$$

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Most popular questions from this chapter

Suppose your graphing calculator has two functions, one called sin \(x,\) which calculates the sine of \(x\) when \(x\) is in radians, and the other called \(s(x),\) which calculates the sine of \(x\) when \(x\) is in degrees. a. Explain why \(s(x)=\sin \left(\frac{\pi}{180} x\right)\) b. Evaluate \(\lim _{x \rightarrow 0} \frac{s(x)}{x} .\) Verify your answer by estimating the limit on your calculator.

Graphing with inverse trigonometric functions a. Graph the function \(f(x)=\frac{\tan ^{-1} x}{x^{2}+1}\) b. Compute and graph \(f^{\prime}\) and determine (perhaps approximately) the points at which \(f^{\prime}(x)=0\) c. Verify that the zeros of \(f^{\prime}\) correspond to points at which \(f\) has a horizontal tangent line.

Find \(f^{\prime}(x), f^{\prime \prime}(x),\) and \(f^{\prime \prime \prime}(x)\) \(f(x)=\frac{1}{x}\)

One of the Leibniz Rules One of several Leibniz Rules in calculus deals with higher-order derivatives of products. Let \((f g)^{(n)}\) denote the \(n\) th derivative of the product \(f g,\) for \(n \geq 1\) a. Prove that \((f g)^{(2)}=f^{\prime \prime} g+2 f^{\prime} g^{\prime}+f g^{\prime \prime}\) b. Prove that, in general,$$(f g)^{(n)}=\sum_{k=0}^{n}\left(\begin{array}{l} n \\\k\end{array}\right) f^{(k)} g^{(n-k)}$$ where \(\left(\begin{array}{l}n \\\ k\end{array}\right)=\frac{n !}{k !(n-k) !}\) are the binomial coefficients. c. Compare the result of (b) to the expansion of \((a+b)^{n}\).

Compute the derivative of the following functions. \(h(x)=\frac{(x+1)}{x^{2} e^{x}}\)
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