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Chapter 3: Problem 15

Evaluate the derivatives of the following functions. $$f(y)=\tan ^{-1}\left(2 y^{2}-4\right)$$

Short Answer

Expert verified
Answer: $$\frac{df}{dy}=\frac{4y}{1+(2y^2-4)^2}$$.

Step by step solution

01

Identify the function and its components

We are given the function $$f(y)=\tan ^{-1}\left(2 y^{2}-4\right)$$. Here, the main function is the inverse tangent function, $$\tan^{-1}$$, and the inner function is $$2y^2 - 4$$.
02

Apply the chain rule

According to the chain rule, the derivative of a composite function can be found by taking the derivative of the outer function and multiplying it by the derivative of the inner function. In our case, the outer function $$g(u)=\tan^{-1}(u)$$ and the inner function $$h(y)=2y^2 - 4$$, so we want to find the derivative of $$g(h(y))$$. So, we have: $$\frac{df}{dy} = \frac{dg}{du} \cdot \frac{dh}{dy}$$
03

Find the derivative of the outer function

The outer function is the inverse tangent function, \(g(u)=\tan^{-1}(u)\). The derivative of the inverse tangent function is: $$\frac{dg}{du} = \frac{1}{1+u^2}$$
04

Find the derivative of the inner function

The inner function is \(h(y) = 2y^2 - 4\), so taking the derivative with respect to y, we have: $$\frac{dh}{dy} = 4y$$
05

Substitute the derivatives back into the chain rule

Now, we substitute the derivatives we found in steps 3 and 4 back into the chain rule: $$\frac{df}{dy} = \frac{dg}{du} \cdot \frac{dh}{dy} = \frac{1}{1+u^2}\cdot 4y$$
06

Replace u with the inner function

Since we know that \(u = 2y^2 - 4\), we can substitute this back into the equation for the derivative: $$\frac{df}{dy}=\frac{1}{1+(2y^2-4)^2}\cdot 4y$$
07

Simplify the expression

The final step is to simplify the expression for the derivative of the function: $$\frac{df}{dy}=\frac{4y}{1+(2y^2-4)^2}$$ The derivative of the function $$f(y)=\tan ^{-1}\left(2 y^{2}-4\right)$$ is $$\frac{df}{dy}=\frac{4y}{1+(2y^2-4)^2}$$.

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Key Concepts

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

Chain Rule
Understanding the chain rule is essential when dealing with composite functions, which are functions that combine two or more functions. In calculus, the chain rule is a formula for computing the derivative of the composition of two or more functions. Let's break it down to make it simple. Imagine you're wearing gloves. To take them off, you first remove the outer glove, and then the inner one. The chain rule works similarly with functions. Starting with the outer function, you differentiate it with respect to an inner function, and then multiply it by the derivative of this inner function itself.

When you need to find the derivative of a function like \( g(h(y)) \), where \( g \) is the outer function and \( h \) is the inner function, the chain rule tells us to calculate it as the derivative of \( g \) with respect to its argument, multiplied by the derivative of \( h \) with respect to \( y \). In terms of our glove analogy, taking off the 'outer glove' is \( g'(h(y)) \), and then removing the 'inner glove' is \( h'(y) \). To summarize, the chain rule formula is expressed as:
Inverse Tangent Derivative
The inverse tangent function, denoted as \( \tan^{-1}(x) \) or \( \arctan(x) \), is a function that returns the angle whose tangent is \( x \). Calculating the derivative of this function is a fundamental concept in calculus, especially when dealing with inverse trigonometric functions. The beauty of the inverse tangent function is its simplicity when differentiated. No matter the complexity of its argument, the derivative of the inverse tangent of \( u \) is \( \frac{1}{1+u^2} \). This simple formula allows us to handle even intricate functions within the inverse tangent.

Not Just a Simple Knack

Students might wonder why the derivative has the form it does. This links back to the geometrical interpretation of the tangent function in relation to the unit circle, and understanding this can enhance the conceptual grasp of why the derivative takes a form that involves squaring the function's argument.
Composite Functions Differentiation
Differentiating composite functions involves using the chain rule, but it's important to conceptualize what a composite function is. A composite function is created when one function is nested inside another. This can be symbolically represented as \( f(g(x)) \), where \( f \) is applied to the result of \( g(x) \). This 'function within a function' setup is prevalent in math problems and real-world applications.

To successfully differentiate a composite function, you must recognize each individual function, find their derivatives separately, and then apply the chain rule to collectively differentiate the entire composition. Keep in mind the order in which functions are layered, as this dictates how the chain rule is applied. Remember, you're peeling away the layers of an onion, from the outermost layer to the inner core. Each step is crucial to ensure accuracy in calculating the derivative of a composite function.

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Most popular questions from this chapter

Economists use production functions to describe how the output of a system varies with respect to another variable such as labor or capital. For example, the production function \(P(L)=200 L+10 L^{2}-L^{3}\) gives the output of a system as a function of the number of laborers \(L\). The average product \(A(L)\) is the average output per laborer when \(L\) laborers are working; that is \(A(L)=P(L) / L\). The marginal product \(M(L)\) is the approximate change in output when one additional laborer is added to \(L\) laborers; that is, \(M(L)=\frac{d P}{d L}\). a. For the given production function, compute and graph \(P, A,\) and \(M\). b. Suppose the peak of the average product curve occurs at \(L=L_{0},\) so that \(A^{\prime}\left(L_{0}\right)=0 .\) Show that for a general production function, \(M\left(L_{0}\right)=A\left(L_{0}\right)\).

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals). A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas \(y=c x^{2}\) form orthogonal trajectories with the family of ellipses \(x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants (see figure). Find \(d y / d x\) for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. \(y=m x ; x^{2}+y^{2}=a^{2},\) where \(m\) and \(a\) are constants

Let \(b\) represent the base diameter of a conifer tree and let \(h\) represent the height of the tree, where \(b\) is measured in centimeters and \(h\) is measured in meters. Assume the height is related to the base diameter by the function \(h=5.67+0.70 b+0.0067 b^{2}\). a. Graph the height function. b. Plot and interpret the meaning of \(\frac{d h}{d b}\).

The bottom of a large theater screen is \(3 \mathrm{ft}\) above your eye level and the top of the screen is \(10 \mathrm{ft}\) above your eye level. Assume you walk away from the screen (perpendicular to the screen) at a rate of \(3 \mathrm{ft} / \mathrm{s}\) while looking at the screen. What is the rate of change of the viewing angle \(\theta\) when you are \(30 \mathrm{ft}\) from the wall on which the screen hangs, assuming the floor is horizontal (see figure)?

Vertical tangent lines a. Determine the points where the curve \(x+y^{2}-y=1\) has a vertical tangent line (see Exercise 53 ). b. Does the curve have any horizontal tangent lines? Explain.
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