Problem 49
Question
In Exercises \(49-70\) , find the derivative of \(y\) with respect to the appropriate variable. $$ y=\cos ^{-1}\left(x^{2}\right) $$
Step-by-Step Solution
Verified Answer
The derivative of \(y\) is \(-\frac{2x}{\sqrt{1-x^4}}\)."
1Step 1: Recognize the Formula for the Derivative of Inverse Cosine
The derivative of the inverse cosine function, \(\cos^{-1}(u)\), with respect to \(u\) is \(-\frac{1}{\sqrt{1-u^2}}\). This will be used once we apply the chain rule to the function. Here, \(u = x^2\).
2Step 2: Apply the Chain Rule
The function \(y = \cos^{-1}(x^2)\) is comprised of an outer function, \(\cos^{-1}(u)\), and an inner function, \(u = x^2\). By the chain rule, the derivative \(\frac{dy}{dx}\) is the derivative of the outer function, evaluated at the inner function, multiplied by the derivative of the inner function.
3Step 3: Derivative of the Outer Function
Using \(u = x^2\), the derivative of the outer function \(\cos^{-1}(u)\) with respect to \(u\) is \(-\frac{1}{\sqrt{1-u^2}}\). Substitute \(u = x^2\) into this formula: \(-\frac{1}{\sqrt{1-(x^2)^2}} = -\frac{1}{\sqrt{1-x^4}}\).
4Step 4: Derivative of the Inner Function
Compute the derivative of the inner function \(u = x^2\) with respect to \(x\), which is \(\frac{du}{dx} = 2x\).
5Step 5: Combine Using the Chain Rule
Using the results from Steps 3 and 4, apply the chain rule: \(\frac{dy}{dx} = -\frac{1}{\sqrt{1-x^4}} \times 2x = -\frac{2x}{\sqrt{1-x^4}}\). This gives the derivative of \(y\) with respect to \(x\).
6Step 6: Verify the Result
Ensure each step has been correctly applied, specifically checking the derivative rules and chain rule application. The chain rule combines the two derivatives into \(-\frac{2x}{\sqrt{1-x^4}}\).
Key Concepts
Inverse Trigonometric FunctionsChain RuleDifferentiation Techniques
Inverse Trigonometric Functions
Inverse trigonometric functions are functions that reverse trigonometric operations, essentially answering the question: what angle creates this trigonometric result? They play a critical role in calculus when dealing with derivatives and integrals because they provide unique solutions for angles given certain values. The inverse cosine function, denoted as \( \cos^{-1}(x) \), is particularly important. It specifies the angle whose cosine equals \(x\). When differentiating inverse trigonometric functions, formulas help guide us. For \( \cos^{-1}(x) \), the derivative formula is
- \( \frac{d}{dx} [\cos^{-1}(x)] = -\frac{1}{\sqrt{1-x^2}} \)
Chain Rule
The chain rule is a fundamental technique in calculus for differentiating composite functions. A composite function arises when one function is nested within another, like \( f(g(x)) \). To differentiate such a composite function, the chain rule states:
- \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
- \( \frac{dy}{dx} = -\frac{1}{\sqrt{1-x^4}} \times 2x = -\frac{2x}{\sqrt{1-x^4}} \)
Differentiation Techniques
Differentiation is the process of finding the rate at which a function changes at any point, known as its derivative. Several techniques can simplify finding derivatives, especially for more complex functions. Key techniques include:
- **Basic Rules**: Power rule, product rule, and quotient rule help with direct functions of \(x\).
- **Chain Rule**: Used, as explained, for composite functions with nested layers.
- **Inverse Function Differentiation**: Specific formulas exist for inverse trigonometric functions, as seen in \( \cos^{-1}(x) \) differentiation.
Other exercises in this chapter
Problem 48
If a composite \(f \circ g\) is one-to-one, must \(g\) be one-to-one? Give reasons for your answer.
View solution Problem 49
Evaluate the integrals in Exercises \(41-50\) . $$ \int \frac{\operatorname{sech} \sqrt{t} \tanh \sqrt{t} d t}{\sqrt{t}} $$
View solution Problem 49
Evaluate the integrals in Exercises \(41-62\). $$ \int \frac{e^{\sqrt{r}}}{\sqrt{r}} d r $$
View solution Problem 49
Evaluate the integrals in Exercises \(37-54\). $$ \int_{0}^{\pi / 2} \tan \frac{x}{2} d x $$
View solution