Problem 16 - Calculus Early Transcendentals

Question

Evaluate the derivatives of the following functions. $$g(z)=\tan ^{-1}(1 / z)$$

Step-by-Step Solution

Answer

Answer: The derivative of the function g(z) with respect to z is \(\frac{-1}{2}\).

Step 1: Identify the outer and inner functions

The function we want to differentiate is g(z) = \(\tan ^{-1}(1 / z)\). The outer function is the inverse tangent function, and the inner function is h(z)=1/z.

Step 2: Find the derivative of the outer function

The derivative of the inverse tangent function is: $$\frac{d}{dz}\tan^{-1}(z) = \frac{1}{1+z^2}$$

Step 3: Find the derivative of the inner function

The inner function is h(z)=1/z. To find its derivative, we can write h(z) as \(z^{-1}\) and apply the power rule: $$\frac{d}{dz}(z^{-1}) = -z^{-2} = -\frac{1}{z^2}$$

Step 4: Apply the chain rule

Now we apply the chain rule to find the derivative of g(z): $$\frac{d}{dz} g(z) = \frac{d}{dz} \tan^{-1}(h(z)) = \frac{d}{dz} \tan^{-1}(1/z)$$ Using the chain rule, we have: $$\frac{d}{dz} g(z) = \frac{1}{1+(1/z)^2} \cdot -\frac{1}{z^2}$$

Step 5: Simplify the expression

Now, we want to simplify our expression. First, replace \((1/z)^2\) with \(1/z^2\) in the denominator of the first fraction: $$\frac{d}{dz} g(z) = \frac{1}{1+1/z^2} \cdot -\frac{1}{z^2}$$ Next, multiply the two fractions: $$\frac{d}{dz} g(z) = \frac{-1}{(1+1/z^2)z^2}$$ Finally, find a common denominator and simplify the expression: $$\frac{d}{dz} g(z) = \frac{-1}{(z^2+z^2)/z^2} = \frac{-1}{(2z^2)/z^2} = \frac{-1}{2}$$ Thus, the derivative of g(z) with respect to z is given by: $$\frac{d}{dz} g(z) = \frac{-1}{2}$$

Key Concepts

Derivative

In calculus, a derivative represents how a function changes as its input changes. In simple terms, a derivative tells us the rate at which one quantity changes with respect to another. It's like measuring how fast you're going by looking at how much distance you cover over a certain time period. For example, if you have a function \( f(x) \), the derivative \( f'(x) \) tells us how \( f(x) \) changes for a tiny change in \( x \).
Let's say we're deciphering the speed of a car by looking at a speedometer: that's effectively what a derivative does for functions. They help in finding slopes of curves, determining maxima and minima, and solving many real-world problems in physics and engineering.
Finding derivatives can often mean using various rules, such as the power rule, product rule, quotient rule, and more. In our exercise, we've used the power rule to find the derivative of the inner function \( h(z) = z^{-1} \). By applying the power rule: \( \frac{d}{dz}(z^{-1}) = -z^{-2} \).

Chain Rule

The chain rule is a fundamental method in calculus for finding the derivative of a composite function. This rule is critical when you have a function nested inside another function. Think of it as peeling back layers of an onion, moving from the outermost layer to the innermost.
In practical terms, if you have a function \( g(z) = \tan^{-1}(1/z) \), it consists of an outer function, \( \tan^{-1} \), and an inner function, \( 1/z \). To differentiate this composite function, you first take the derivative of the outer function, then multiply by the derivative of the inner one.

• First, differentiate \( \tan^{-1} \), which gives: \( \frac{1}{1+(1/z)^2} \).
• Then multiply it by the derivative of the inner function, \(-1/z^{2} \).

This combination gives us the derivative of the composite function, illustrating how the chain rule handles these layered situations seamlessly.

Inverse Trigonometric Functions

Inverse trigonometric functions are the opposite of regular trigonometric functions like sine, cosine, and tangent. They help find angles when the values of the trigonometric ratios are known. These functions include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \), among others.
For example, \( \tan^{-1}(x) \) returns the angle whose tangent is \( x \). These functions are essential when dealing with right triangles and also in calculus for integrating and differentiating inverse relationships.
In our exercise, \( \tan^{-1}(1/z) \) poses a unique function where we have to deal with differentiating an inverse trigonometric function soundly. The derivative of \( \tan^{-1}(z) \) is \( \frac{1}{1+z^2} \), a formula used extensively whenever you encounter this inverse trig function.
By knowing these derivatives, we can handle complex expressions and solve equations involving inverse trigonometric terms effectively.