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Find \(f^{\prime}(x)\) $$f(x)=\tan ^{4}\left(x^{3}\right)$$

Short Answer

Expert verified
The derivative is \( f'(x) = 12x^2 \cdot (\tan(x^3))^{3} \cdot \sec^2(x^3) \).

Step by step solution

01

Recognize the Power Rule

Notice that the function is in the form of a power function with the power of 4 applied to the inner function. We apply the power rule for derivatives: if \( y = u^n \), then \( \frac{dy}{dx} = n \cdot u^{n-1} \cdot \frac{du}{dx} \). Here, \( u = \tan(x^3) \) and \( n = 4 \).
02

Differentiate the Outer Function

Apply the power rule to differentiate the outer function: \( (\tan(x^3))^4 \). This gives: \( 4 \cdot (\tan(x^3))^{3} \cdot \frac{d}{dx}[\tan(x^3)] \).
03

Differentiate the Inner Function Using Chain Rule

Differentiate \( \tan(x^3) \) using the chain rule. First, note that the derivative of \( \tan(u) \) is \( \sec^2(u) \). Here, \( u = x^3 \), so the derivative of the \( \tan \) part is \( \sec^2(x^3) \cdot \frac{d}{dx}[x^3] \).
04

Differentiate the Innermost Function

Differentiate \( x^3 \) to get \( 3x^2 \). Thus, \( \frac{d}{dx}[x^3] = 3x^2 \).
05

Compile the Derivatives Together

Substitute the derived components back into the expression. The derivative is thus: \[ f'(x) = 4 \cdot (\tan(x^3))^{3} \cdot \sec^2(x^3) \cdot 3x^2 \].
06

Simplify the Expression

Simplify the result to get: \[ f'(x) = 12x^2 \cdot (\tan(x^3))^{3} \cdot \sec^2(x^3) \].

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Key Concepts

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

Chain Rule in Calculus
The chain rule is a fundamental principle in calculus that's essential when dealing with composite functions. Think of a composite function as a function inside another function. This rule lets us find the derivative of such complex setups by breaking them down into smaller, more manageable parts.
Imagine you have a function like \( g(h(x)) \). Here, \( h(x) \) is the inner function and \( g(u) \) is the outer function, with \( u = h(x) \). The chain rule states that the derivative of this composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function:
  • If \( y = g(h(x)) \), then \( \frac{dy}{dx} = g'(h(x)) \cdot h'(x) \).
In our specific problem, we have \( f(x) = \tan^4(x^3) \). We first differentiate the outer part, \( \tan^4(u) \), using the power rule (explained further below). Then, we handle the inner function, \( \tan(x^3) \), by recognizing its derivative as \( \sec^2(u) \). Finally, the very inner part \( x^3 \) is differentiated using basic rules, resulting in \( 3x^2 \).By connecting these steps with the chain rule, we achieve the complete derivative of complex nested functions.
Power Rule Differentiation
The power rule is a straightforward yet powerful tool for differentiation. It applies to functions that are expressed in terms of powers of a variable. This rule states that if you have a function of the form \( y = x^n \), then the derivative \( y' \) is given by:
  • \( \frac{dy}{dx} = n \cdot x^{n-1} \).
The beauty of the power rule is its simplicity and versatility, making it one of the first rules students learn in calculus.
When applied in our exercise, the power rule is instrumental in taking the derivative of the outer function \( (\tan(x^3))^4 \). By setting the inner function as \( u = \tan(x^3) \), the power rule helps to initially differentiate \( u^4 \). This results in \( 4 \cdot u^{3} \cdot \frac{du}{dx} \), which sets the stage for implementing the chain rule correctly.
The power rule is particularly helpful here as it simplifies the complex nested function into a manageable form, allowing us to focus next on the inner derivatives, smoothly connecting into the chain rule process.
Trigonometric Derivatives
Trigonometric derivatives deal with finding the rate of change of trigonometric functions. They play a crucial role when these functions appear in calculus problems. Among them, the derivative of \( \tan(x) \) is one of the critical calculations:
  • \( \frac{d}{dx}[\tan(x)] = \sec^2(x) \).
This derivative emerges frequently, especially when trigonometric functions are within composite functions like in our example with \( \tan(x^3) \).
Understanding trigonometric derivatives allows us to extend these rules to more complex scenarios, exactly like the chain rule necessitates. In the exercise at hand, \( \tan(x^3) \) must be differentiated, which means we use the derivative of \( \tan(x) \), yielding \( \sec^2(x^3) \).
Additionally, we also need to multiply by the derivative of the innermost function \( x^3 \), which simplifies to \( 3x^2 \). This process emphasizes the importance of connecting basic trigonometric rules with more intricate derivatives.
Mastering trigonometric derivatives provides the tools needed to tackle broader applications within calculus, ensuring a well-rounded grasp of calculus concepts.

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Most popular questions from this chapter

In each part, compute \(f^{\prime}, f^{\prime \prime}, f^{\prime \prime \prime},\) and then state the formula for \(f^{(n)}\) (a) \(f(x)=1 / x\) (b) \(f(x)=1 / x^{2}\) IHint: The expression \((-1)^{n}\) has a value of 1 if \(n\) is even and -1 if \(n\) is odd. Use this expression in your answer. 1

(a) Use a graphing utility to obtain the graph of the function \(f(x)=\sin x^{2} \cos x\) over the interval \([-\pi / 2, \pi / 2]\) (b) Use the graph in part (a) to make a rough sketch of the graph of \(f^{\prime}\) over the interval. (c) Find \(f^{\prime}(x)\), and then check your work in part (b) by using the graphing utility to obtain the graph of \(f^{\prime}\) over the interval. (d) Find the equation of the tangent line to the graph of \(f\) at \(x=1,\) and graph \(f\) and the tangent line together over the interval.

Find \(f^{\prime}(x)\) $$f(x)=\left[x^{4}-\sec \left(4 x^{2}-2\right)\right]^{-4}$$

True-False Determine whether the statement is true or false. Explain your answer. The velocity of an object represents a change in the object's position.

A function \(y=f(x)\) and values of \(x_{0}\) and \(x_{1}\) are given. (a) Find the average rate of change of \(y\) with respect to \(x\) over the interval \(\left[x_{0}, x_{1}\right]\). (b) Find the instantaneous rate of change of \(y\) with respect to \(x\) at the specified value of \(x_{0}\). (c) Find the instantaneous rate of change of \(y\) with respect to \(x\) at an arbitrary value of \(x_{0}\). (d) The average rate of change in part (a) is the slope of a certain secant line, and the instantaneous rate of change in part (b) is the slope of a certain tangent line. Sketch the graph of \(y=f(x)\) together with those two lines. $$y=x^{3} ; x_{0}=1, x_{1}=2$$

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