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Chapter 3: Problem 15

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

Short Answer

Expert verified
Question: Find the derivative of the function \(f(y) = \tan^{-1}(2y^2-4)\). Answer: The derivative of the given function is \(f'(y) = \frac{4y}{1+(2y^2-4)^2}\).

Step by step solution

01

Differentiate the outer function

First, we differentiate the outer function \(\tan^{-1}(u)\) with respect to \(u\). The derivative of the inverse tangent function is: $$\frac{d}{du}\left( \tan^{-1}(u) \right) = \frac{1}{1+u^2}$$
02

Differentiate the inner function

Now, we differentiate the inner function \(u(y)=2y^2-4\) with respect to \(y\). This is a simple polynomial, so using the power rule we get: $$\frac{du}{dy}\left(2y^2-4\right) = 4y$$
03

Apply the chain rule

Now that we have both derivatives, we can apply the chain rule by multiplying them: $$\frac{d}{dy}f(y) = \frac{d}{du}\left(\tan^{-1}(u(y))\right) \cdot \frac{du}{dy}\left(2y^2-4\right)$$ Substitute the derivatives of the outer and inner functions from Steps 1 and 2: $$\frac{d}{dy}f(y) = \frac{1}{1+(2y^2-4)^2} \cdot 4y$$
04

Simplify the expression

Finally, simplify the expression to obtain the derivative of the given function: $$\frac{d}{dy}f(y) = \frac{4y}{1+(2y^2-4)^2}$$ So, the derivative of the function \(f(y) = \tan^{-1}(2y^2-4)\) is: $$f'(y) = \frac{4y}{1+(2y^2-4)^2}$$

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Key Concepts

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

Chain Rule
The chain rule is crucial when dealing with complex functions, especially those where a function is nested within another. For example, in evaluating the derivative of \( f(y) = \tan^{-1}(2y^2 - 4) \), we use the chain rule because it involves the composition of two functions. Here is how it works:
  • Identify the outer function: In this case, it is \( \tan^{-1}(u) \), where \( u = 2y^2 - 4 \).
  • Identify the inner function: It is \( u(y) = 2y^2 - 4 \).
  • Differentiate each function separately: First take the derivative of the outer function with respect to \( u \) and then of the inner function with respect to \( y \).
  • Finally, multiply these derivatives: This gives you the derivative of the original composite function.
This technique is invaluable when working with nested functions as it makes the differentiation process more structured and manageable.
Inverse Trigonometric Functions
Inverse trigonometric functions like \( \tan^{-1}(u) \) are often involved in calculus problems. They can be tricky because their derivatives are not as simple as basic trigonometric functions. Let's break it down:
  • Derivative of \( \tan^{-1}(u) \) is \( \frac{1}{1+u^2} \).
  • These functions are used to find angles when given a ratio.
  • In our example, \( \, f(y) = \tan^{-1}(2y^2 - 4) \, \) requires understanding how the angle changes as the value \( 2y^2 - 4 \) changes.
Inverse trigonometric functions have restrictions on their domains because they otherwise don't have a unique inverse. Knowing these nuances is key to successfully navigating calculus involving these functions. When differentiating, always remember the derivative formulas specific to each inverse trigonometric function as they aid in solving calculus problems correctly.
Polynomial Differentiation
Polynomial differentiation involves finding the derivative of polynomial expressions, which are sums of constant coefficients multiplying variables raised to a power. In our example, the inner function \( u(y) = 2y^2 - 4 \) is a simple polynomial. Here's a step-by-step guide to differentiating it:
  • Identify each term in the polynomial.
  • Apply the power rule: For a term \( ay^n \), the derivative is \( nay^{n-1} \).
  • Differentiate each term separately: For \( 2y^2 \), this becomes \( 4y \).
  • Combine differentiated terms: In this instance, only one non-constant term is differentiated, leaving the constant subtracted as zero.
Polynomial differentiation is straightforward because it follows a consistent rule set, making it an approachable yet powerful tool for finding the rate of change in algebraic expressions. Remember, understanding these basic techniques simplifies tackling more complex calculus problems.

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Most popular questions from this chapter

Gravitational force The magnitude of the gravitational force between two objects of mass \(M\) and \(m\) is given by \(F(x)=-\frac{G M m}{x^{2}},\) where \(x\) is the distance between the centers of mass of the objects and \(G=6.7 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\) is the gravitational constant (N stands for newton, the unit of force; the negative sign indicates an attractive force). a. Find the instantaneous rate of change of the force with respect to the distance between the objects. b. For two identical objects of mass \(M=m=0.1 \mathrm{kg},\) what is the instantaneous rate of change of the force at a separation of \(x=0.01 \mathrm{m} ?\) c. Does the instantaneous rate of change of the force increase or decrease with the separation? Explain.

Use any method to evaluate the derivative of the following functions. $$h(x)=\left(5 x^{7}+5 x\right)\left(6 x^{3}+3 x^{2}+3\right)$$

Visualizing tangent and normal lines a. Determine an equation of the tangent line and normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises \(63-68 .)\) b. Graph the tangent and normal lines on the given graph. $$\begin{aligned}&3 x^{3}+7 y^{3}=10 y\\\&\left(x_{0}, y_{0}\right)=(1,1)\end{aligned}$$

Suppose \(y=L(x)=a x+b\) (with \(a \neq 0\) ) is the equation of the line tangent to the graph of a one-to-one function \(f\) at \(\left(x_{0}, y_{0}\right) .\) Also, suppose that \(y=M(x)=c x+d\) is the equation of the line tangent to the graph of \(f^{-1}\) at \(\left(y_{0}, x_{0}\right)\) a. Express \(a\) and \(b\) in terms of \(x_{0}\) and \(y_{0}\) b. Express \(c\) in terms of \(a,\) and \(d\) in terms of \(a, x_{0},\) and \(y_{0}\) c. Prove that \(L^{-1}(x)=M(x)\)

Cobb-Douglas production function The output of an economic system \(Q,\) subject to two inputs, such as labor \(L\) and capital \(K\) is often modeled by the Cobb- Douglas production function \(Q=c L^{a} K^{b} .\) When \(a+b=1,\) the case is called constant returns to scale. Suppose \(Q=1280, a=\frac{1}{3}, b=\frac{2}{3},\) and \(c=40\) a. Find the rate of change of capital with respect to labor, \(d K / d L\) b. Evaluate the derivative in part (a) with \(L=8\) and \(K=64\)
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