Problem 35 - Calculus of a Single Variable

Question

In Exercises 33–36, find an equation of the tangent line to the graph of the function at the given point. $$ y=(\cosh x-\sinh x)^{2}, \quad(0,1) $$

Step-by-Step Solution

Answer

The equation of the tangent line to the graph of \(y=(\cosh x-\sinh x)^2\) at the point \((0,1)\) is \(y = 2x + 1\).

Step 1: Compute the derivative

First, let's find the derivative of \(y=(\cosh x-\sinh x)^2\). Using the Chain Rule, we have \[y' = 2(\cosh x-\sinh x)(\cosh x + \sinh x)\] Since, \(\cosh^2 x - \sinh^2 x = 1\), the derivative simplifies to \[y' = 2\]

Step 2: Find the slope of the tangent line at the given point

Substitute \(x = 0\) into \(y'\) to find the slope at point \((0,1)\). Doing so gives us \(m = 2\)

Step 3: Write the equation of the tangent line

Now, using the slope \(m\) and the point \((0,1)\) in the point-slope form of the line equation, we get \[y - 1 = 2(x - 0)\] Simplifying, we get the equation of the tangent line: \[y = 2x + 1\]

Key Concepts

Calculus

Calculus is a vast branch of mathematics that deals with change and motion. In the challenge of understanding geometric shapes, physical phenomena, and much more, calculus offers tools like derivatives and integrals to dissect and analyze variable quantities.

When faced with a function, like the one in the exercise \(y=(\cosh x-\sinh x)^{2}\), calculus provides the methods to find the rate of change at a particular point, which is exactly what's needed to determine the equation of a tangent line. It's the study of how things change and the rates at which they change that lies at the core of calculus and is essential for finding the tangent line to a curve at a given point.

Derivative

The derivative is a fundamental concept within calculus, encapsulating the idea of an instantaneous rate of change. It can be thought of as the slope of the tangent line to the function at any point. In the given exercise, we seek the derivative of \(y=(\cosh x-\sinh x)^{2}\) because it tells us how steep the tangent line is at the point \(x=0\).

The derivative is not just a single step operation; it's the culmination of understanding limits and applying rules to find the rate at which a function is changing at a particular point. Once the derivation is done, as in the provided solution \(y'=2\), we possess the vital information to construct the equation of the tangent line.

Chain Rule

The Chain Rule is a critical tool in calculus used for computing the derivative of a composite function. Essentially, when a function is composed of an outer and an inner function, the derivative involves taking the derivatives of both, and 'chaining' them together.

For the equation \(y=(\cosh x-\sinh x)^{2}\), we identify \(u=\cosh x-\sinh x\) as the inner function and \(u^2\) as the outer function. By applying the Chain Rule \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\), we find the derivative of the outer function with respect to the inner, \((\frac{du^2}{du}\)), and multiply it by the derivative of the inner function with respect to \(x\). The solution simplifies the derivative using the identity \(\cosh^2 x - \sinh^2 x = 1\), demonstrating a clever use of the Chain Rule in conjunction with hyperbolic identities.

Hyperbolic Functions

Hyperbolic functions, including \(\cosh x\) and \(\sinh x\), are analogs of the trigonometric functions but for a hyperbola rather than a circle. These functions show up in numerous areas, from architecture to theoretical physics, and are essential for solving various types of equations.

In our context, understanding how \(\cosh\) and \(\sinh\) derivatives behave is key to finding the slope of our tangent line. The exercise takes advantage of a pivotal hyperbolic identity \(\cosh^2 x - \sinh^2 x = 1\) to simplify the derivative. Recognizing and applying such identities is crucial when working with hyperbolic functions to derive simplified forms or solve complex equations.