Problem 52 - Calculus Early Transcendentals

Question

Evaluate the following derivatives. \(f(u)=\sinh ^{-1}(\tan u)\)

Step-by-Step Solution

Answer

Question: Find the derivative of the function \(f(u) = \sinh^{-1}(\tan{u})\). Answer: The derivative of the given function is \(\frac{d}{du} f(u) = \frac{\sec^2(u)}{\sqrt{1+\tan^2(u)}}\).

Step 1: Identify the outer and inner functions

The outer function is the inverse hyperbolic sine function, denoted by \(\sinh^{-1}(x)\). The inner function is the tangent function, denoted by \(\tan(u)\).

Step 2: Find the derivative of the outer function

The derivative of the inverse hyperbolic sine function is given by: \(\frac{d}{dx} \sinh^{-1}(x) = \frac{1}{\sqrt{1+x^2}}\).

Step 3: Find the derivative of the inner function

The derivative of the tangent function is given by: \(\frac{d}{du} \tan(u) = \sec^2(u)\).

Step 4: Apply the chain rule

Now, apply the chain rule to find the derivative of the composite function: \(\frac{d}{du} f(u) = \frac{d}{du} \sinh^{-1}(\tan{u}) = \frac{d}{dx} \sinh^{-1}(x) \cdot \frac{d}{du} \tan(u)\) where \(x = \tan(u)\).

Step 5: Calculate the derivative of the entire function

Substitute the derivatives of the outer and inner functions from Steps 2 and 3, as well as the expression for \(x\), into the chain rule formula: \(\frac{d}{du} f(u) = \frac{1}{\sqrt{1+\tan^2(u)}} \cdot \sec^2(u)\). The derivative of the given function is: \(\frac{d}{du} f(u) = \frac{\sec^2(u)}{\sqrt{1+\tan^2(u)}}\).

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus that help us understand how a function changes at any given point. They provide the slope of the tangent line to the curve of a function, essentially telling us about rates of change.

• The basic idea behind finding a derivative is to find the limit of the average rate of change of the function as the interval approaches zero.
• If you imagine a curve, the derivative at a point gives you the steepness and direction of the curve at that point.

Understanding derivatives is crucial because they are used in a variety of real-world applications, from calculating velocity in physics to finding maximum profit in business.
In our exercise, we are finding the derivative of a composite function which involves an inverse hyperbolic function and a trigonometric function. This requires applying some special rules like the chain rule, which we'll explore next.

Chain Rule

The chain rule is a powerful tool in calculus that allows us to differentiate composite functions. A composite function is made up of two or more functions, where one function is nested inside another.

• The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, times the derivative of the inner function.
• Mathematically, if we have a function of the form \(y = g(f(x))\), the derivative can be expressed as \(\frac{dy}{dx} = g'(f(x)) \cdot f'(x)\).

Applying the chain rule can make the process of differentiation much simpler when dealing with functions like \(f(u)=\sinh^{-1}(\tan u)\).
In this exercise, the outer function is the inverse hyperbolic sine and the inner function is the tangent of \(u\). By following the chain rule, we are able to systematically find the derivative of the entire function, even when it involves complex combinations.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are analogs of the inverse trigonometric functions but for hyperbolic functions. These functions include \(\sinh^{-1}(x)\), \(\cosh^{-1}(x)\), and \(\tanh^{-1}(x)\) among others.

• The inverse hyperbolic sine function, \(\sinh^{-1}(x)\), is particularly interesting because it can be expressed with a logarithmic form: \(\sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1})\).
• Its derivative is \(\frac{d}{dx} \sinh^{-1}(x) = \frac{1}{\sqrt{1+x^2}}\), which reveals that the growth rate slows down as \(x\) becomes larger.

In our problem, understanding the derivative of \(\sinh^{-1}(x)\) is crucial because it forms the outer part of our composite function.
Handling such functions requires a solid grasp of both hyperbolic identities and differential calculus, by which we unfold intricate problems into simpler parts using derivative rules.