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Chapter 3: Problem 21

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\cos ^{-1}\left(x^{2}\right)$$

Short Answer

Expert verified
The derivative is \( \frac{-2x}{\sqrt{1-x^4}} \).

Step by step solution

01

Identify the Outer Function

The given function is a composite function where the outer function is the inverse cosine function, which is written as \( \cos^{-1}(u) \), where \( u = x^2 \).
02

Differentiate the Outer Function

The derivative of \( \cos^{-1}(u) \) with respect to \( u \) is \( \frac{-1}{\sqrt{1-u^2}} \). So, when \( u = x^2 \), the derivative is \( \frac{-1}{\sqrt{1-(x^2)^2}} \).
03

Differentiate the Inner Function

The inner function, \( u = x^2 \), has the derivative \( \frac{du}{dx} = 2x \).
04

Apply the Chain Rule

According to the chain rule, the derivative of a composite function \( y = f(g(x)) \) is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). Here, substitute the derivatives found: \( \frac{-1}{\sqrt{1-(x^2)^2}} \cdot 2x \).
05

Simplify the Derivative Expression

Simplify the expression: \( \frac{dy}{dx} = \frac{-2x}{\sqrt{1-x^4}} \). Thus, the derivative of \( y \) with respect to \( x \) is \( \frac{-2x}{\sqrt{1-x^4}} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Trigonometric Functions
Inverse trigonometric functions are crucial when dealing with angles and their relations to lengths in triangles. They help us to "undo" the trigonometric functions and find angles based on ratios. In our exercise, \(\cos^{-1}(x^2)\)is an inverse cosine function. The function's output is an angle, whose cosine is \(x^2\). Recognizing the nature of inverse trigonometric functions is key when differentiating, as their derivatives have special forms. The derivative of \(\cos^{-1}(u)\) based on \(u\) is \\(\frac{-1}{\sqrt{1-u^2}}\). This expression accounts for the rate of change of the angle concerning changes in its cosine angle.
Composite Functions
A composite function is essentially a function within another function. In simpler terms, you take one function's output and use it as the input to another. The exercise gives us \(\cos^{-1}(x^2)\), where \(\cos^{-1}\) is the outer function, and \(x^2\) is the inner function. Understanding composite functions involves recognizing these layers. First, deal with the outermost function, then work inward to the core. Recognizing such structures is essential in calculus, especially when differentiating, as it determines how you apply rules like the chain rule.
Chain Rule
The chain rule is a fundamental method for finding derivatives, particularly with composite functions. It allows us to differentiate "layer by layer." According to the chain rule, if you have \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\). To apply this to our exercise, we first differentiate the outer function, the inverse cosine, and then differentiate the inner function, \(x^2\). \[ f'(g(x)) = \frac{-1}{\sqrt{1-(x^2)^2}}, \quad g'(x) = 2x. \] Finally, multiply the derivatives of both functions: \(\frac{-1}{\sqrt{1-(x^2)^2}} \cdot 2x\). The chain rule provides a systematic way to tackle complex derivatives by breaking them into manageable parts.
Derivative Simplification
Once you have applied the chain rule, it's common to end up with complex expressions. Therefore, simplification is essential. In our derivative calculation of \(\cos^{-1}(x^2)\), we found the raw derivative as \(\frac{-1}{\sqrt{1-(x^2)^2}} \cdot 2x\). The next step is to simplify: \[ \text{Simplified Expression: } \frac{-2x}{\sqrt{1-x^4}}. \] Simplifying involves arithmetic operations, algebra, and sometimes trigonometric identities. Here, simplifying helps in making the derivative easier to interpret and more practical for further mathematical operations or problem-solving. A clear, concise derivative is always preferable, aiding both comprehension and utility.

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Most popular questions from this chapter

Find the derivative of \(y\) with respect to the given independent variable. $$y=\log _{2}\left(\frac{x^{2} e^{2}}{2 \sqrt{x+1}}\right)$$

If \(f(x)=x^{n}, n \geq 1,\) show from the definition of the derivative that \(f^{\prime}(0)=0\)

Slopes on sine curves a. Find equations for the tangent lines to the curves \(y=\sin 2 x\) and \(y=-\sin (x / 2)\) at the origin. Is there anything special about how the tangent lines are related? Give reasons for your answer. b. Can anything be said about the tangent lines to the curves \(y=\sin m x\) and \(y=-\sin (x / m)\) at the origin (m a constant \(\neq 0\) )? Give reasons for your answer. c. For a given \(m,\) what are the largest values the slopes of the curves \(y=\sin m x\) and \(y=-\sin (x / m)\) can ever have? Give reasons for your answer. d. The function \(y=\sin x\) completes one period on the interval \([0,2 \pi],\) the function \(y=\sin 2 x\) completes two periods, the function \(y=\sin (x / 2)\) completes half a period, and so on. Is there any relation between the number of periods \(y=\sin m x\) completes on \([0,2 \pi]\) and the slope of the curve \(y=\sin m x\) at the origin? Give reasons for your answer.

In Exercises \(69-74,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\). c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can that satisfies \(|x-a|<\delta \Rightarrow|f(x)-L(x)|<\varepsilon\) for \(\varepsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see whether your \(\delta\) -estimate holds true. $$f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2$$

In Exercises \(69-74,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\). c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can that satisfies \(|x-a|<\delta \Rightarrow|f(x)-L(x)|<\varepsilon\) for \(\varepsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see whether your \(\delta\) -estimate holds true. $$f(x)=\sqrt{x} \sin ^{-1} x,[0,1], \quad a=\frac{1}{2}$$
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