Problem 17
Question
In Exercises \(9-22,\) write the function in the form \(y=f(u)\) and \(u=g(x) .\) Then find \(d y / d x\) as a function of \(x\). $$y=\tan ^{3} x$$
Step-by-Step Solution
Verified Answer
\( \frac{dy}{dx} = 3\tan^2 x \sec^2 x \).
1Step 1: Rewrite the Function
To express the function in the form \(y = f(u)\) where \(u = g(x)\), identify the inner function and the outer function. For \(y = \tan^3 x\), the function inside the cube is the tangent function. So, let \(u = g(x) = \tan x\). Now, rewrite the function as \(y = f(u) = u^3\).
2Step 2: Find the Derivative of f with respect to u
Now, differentiate the function \(y = u^3\) with respect to \(u\). The derivative \( \frac{d}{du}(u^3) = 3u^2 \). Thus, we have \( \frac{dy}{du} = 3u^2 \).
3Step 3: Differentiate g with respect to x
Differentiate \(u = \tan x\) with respect to \(x\). The derivative of \(\tan x\) is \( \sec^2 x \). Therefore, \( \frac{du}{dx} = \sec^2 x \).
4Step 4: Apply the Chain Rule
To find \( \frac{dy}{dx} \), apply the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). We have already found \( \frac{dy}{du} = 3u^2 \) and \( \frac{du}{dx} = \sec^2 x \). So, \( \frac{dy}{dx} = 3u^2 \times \sec^2 x \).
5Step 5: Substitute Back for u
Now substitute back \( u = \tan x\) into the expression for \( \frac{dy}{dx} \). So, \( \frac{dy}{dx} = 3(\tan x)^2 \sec^2 x \). Simplifying slightly gives \( \frac{dy}{dx} = 3\tan^2 x \sec^2 x \).
Key Concepts
Chain RuleTrigonometric FunctionsDerivatives
Chain Rule
The chain rule is a key concept in calculus that allows us to differentiate composite functions. Think of composite functions as layers of rules applied to an original variable. For example, if you have a function of the form \(y = f(g(x))\), it consists of an outer function \(f\) and an inner function \(g\).
To find the derivative of \(y\) with respect to \(x\), we need to use the chain rule, which states that:
This method is particularly handy when dealing with functions that can be split into simpler components for differentiation. It effectively "chains" the derivatives of these components together. In our exercise, this rule was used to find the derivative of \(y = \tan^3{x}\). By recognizing \(u = \tan{x}\) and \(f(u) = u^3\), we simplified the differentiation process using the chain rule.
To find the derivative of \(y\) with respect to \(x\), we need to use the chain rule, which states that:
- First, differentiate \(f\) with respect to \(u\), which gives \(\frac{dy}{du}\).
- Then, differentiate \(g(x)\) with respect to \(x\), giving \(\frac{du}{dx}\).
- Finally, multiply these derivatives together: \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).
This method is particularly handy when dealing with functions that can be split into simpler components for differentiation. It effectively "chains" the derivatives of these components together. In our exercise, this rule was used to find the derivative of \(y = \tan^3{x}\). By recognizing \(u = \tan{x}\) and \(f(u) = u^3\), we simplified the differentiation process using the chain rule.
Trigonometric Functions
Trigonometric functions are fundamental in calculus because they often appear in various kinds of mathematical problems. Known for their periodic nature, these functions include sine, cosine, and tangent among others.
Tangent function, denoted as \(\tan{x}\), plays a pivotal role in calculus and trigonometry. It's defined as the ratio of the sine and cosine functions: \(\tan{x} = \frac{\sin{x}}{\cos{x}}\).
In the exercise, we encountered the function \(y = \tan^3{x}\), which involves raising the tangent function to the power of three. In this context, the derivative of \(\tan{x}\) with respect to \(x\) is \(\sec^2{x}\), where \(\sec{x} = \frac{1}{\cos{x}}\).
This derivative is crucial in the chain rule process to derive functions that are composed with trigonometric components. Understanding how trigonometric identities and derivatives work provides a solid foundation when dealing with more complex expressions.
Tangent function, denoted as \(\tan{x}\), plays a pivotal role in calculus and trigonometry. It's defined as the ratio of the sine and cosine functions: \(\tan{x} = \frac{\sin{x}}{\cos{x}}\).
In the exercise, we encountered the function \(y = \tan^3{x}\), which involves raising the tangent function to the power of three. In this context, the derivative of \(\tan{x}\) with respect to \(x\) is \(\sec^2{x}\), where \(\sec{x} = \frac{1}{\cos{x}}\).
This derivative is crucial in the chain rule process to derive functions that are composed with trigonometric components. Understanding how trigonometric identities and derivatives work provides a solid foundation when dealing with more complex expressions.
Derivatives
Derivatives are a central concept in calculus and essentially measure how a function changes as its input changes. It can be thought of as the "slope" of the function at a given point.
When working through the exercise, derivatives form the backbone of each step. First, we needed to find the derivative of the outer function \(f(u) = u^3\), resulting in \(\frac{dy}{du} = 3u^2\). Then, recognizing \(u = \tan{x}\), we found \(\frac{du}{dx} = \sec^2{x}\).
The derivative \(\frac{dy}{dx}\) represents how the entire expression \(y = \tan^3{x}\) changes with respect to \(x\).
Substituting back the expression for \(u\) at the end allowed us to arrive at the final derivative: \(\frac{dy}{dx} = 3\tan^2{x} \cdot \sec^2{x}\). Understanding this process shows how different derivatives come together to solve complex problems in an organized manner.
When working through the exercise, derivatives form the backbone of each step. First, we needed to find the derivative of the outer function \(f(u) = u^3\), resulting in \(\frac{dy}{du} = 3u^2\). Then, recognizing \(u = \tan{x}\), we found \(\frac{du}{dx} = \sec^2{x}\).
The derivative \(\frac{dy}{dx}\) represents how the entire expression \(y = \tan^3{x}\) changes with respect to \(x\).
Substituting back the expression for \(u\) at the end allowed us to arrive at the final derivative: \(\frac{dy}{dx} = 3\tan^2{x} \cdot \sec^2{x}\). Understanding this process shows how different derivatives come together to solve complex problems in an organized manner.
Other exercises in this chapter
Problem 17
Faster than a calculator Use the approximation \((1+x)^{k} \approx\) \(1+k x\) to estimate the following. a. \((1.0002)^{50}\) b. \(\sqrt[3]{1.009}\)
View solution Problem 17
Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\ln (\theta+1)-e^{\theta}$$
View solution Problem 17
Find \(d r / d \theta\) $$\theta^{1 / 2}+r^{1 / 2}=1$$
View solution Problem 17
Find \(d y / d x\). $$f(x)=x^{3} \sin x \cos x$$
View solution