Problem 36 - 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. $$ y=e^{\sinh x}, \quad(0,1) $$

Step-by-Step Solution

Answer

The equation of the tangent line to the function \(y = e^{\sinh x}\) at the point (0,1) is \(y = x + 1\).

Step 1: Calculate the derivative

Find the derivative of the function \(y = e^{\sinh x}\) using the chain rule. The derivative of the function \(y = e^{u}\) is \(dy/dx = e^u \cdot du/dx\), where \(u = \sinh x\). Hence \[y' = e^{\sinh x} \cdot \cosh x\].

Step 2: Evaluate the derivative at the given point

Now we need to evaluate the derivative at the given point (0,1). Remember, the derivative gives the slope of the tangent line at any given point. When we substitute x=0 into the derivative, we get \[y'(0) = e^{\sinh(0)} \cdot \cosh(0) = 1\cdot 1 = 1\]. Thus, the slope of the tangent line at the point (0,1) is 1.

Step 3: Find the equation of the tangent line

To find the equation of the tangent line, we will use the point-slope form of a line, which is \(y - y_1 = m(x - x_1)\). Substitute \(m = 1\) from step 2 and point (0,1) into this formula. Our equation hence becomes \(y - 1 = 1*(x - 0) = x\). Simplifying gives us the final equation of the tangent line: \(y = x + 1\).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus that helps us find the derivative of composite functions. If you have a function that is made up of two or more functions, the chain rule is your go-to tool. For instance, if we consider a function in the form of \( y = f(g(x)) \), the derivative using the chain rule is \( dy/dx = f'(g(x)) \cdot g'(x) \). By breaking down the components, we can find how quickly each part of the function changes.

In the given exercise, where the function is \( y = e^{\sinh x} \), we apply the chain rule to this composite structure. The outer function here is \( e^u \) and the inner function is \( u = \sinh x \). The derivative of the outer function is \( e^u \) and that of the inner function \( \sinh x \) is \( \cosh x \).

Putting it all together using the chain rule gives us the derivative: \[ y' = e^{\sinh x} \cdot \cosh x \].

• Composite Functions: Functions within functions.
• Inner and Outer Functions: Distinguishing parts of the function to apply the chain rule.

Remember, the chain rule allows us to peel back the layers of functions and differentiate step by step.

Derivative

In calculus, the derivative is crucial for understanding how a function behaves. It's essentially a tool to determine the rate at which a function's value changes as its input changes. To put it simply, derivatives provide us with the slope of the tangent line to the function at any specific point.

In this exercise, we were tasked with finding the derivative of the function \( y = e^{\sinh x} \) at the point \( (0,1) \). This means we needed to compute \( dy/dx \), which we accomplished using the chain rule as seen in the first section.

Substituting \( x = 0 \) into our derived expression \( y' = e^{\sinh x} \cdot \cosh x \), we found:

• \( \sinh(0) = 0 \)
• \( e^0 = 1 \)

Thus, the expression simplifies to \( y'(0) = 1 \cdot 1 = 1 \), meaning the slope of the tangent line at the point \( (0,1) \) is 1.
This slope tells us that at the point \( (0,1) \), the function is increasing at a rate of 1 unit up for every 1 unit it moves along the x-axis.

Point-Slope Form

The point-slope form is a nifty formula in algebra used to define the equation of a straight line. It's particularly handy when you know a point on the line and its slope. The formula looks like \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the given point on the line.

Having already found the slope \( m = 1 \) from our derivative calculations at point \( (0,1) \), we can plug these into the point-slope form. Here's the breakdown:

• Given point: \( (0, 1) \)
• Slope \( m = 1 \)

Substitute into the formula:

\[ y - 1 = 1(x - 0) \]

Simplifying this gives us the equation of the tangent line:
\[ y = x + 1 \]
The equation tells us how the tangent behaves at point \( (0,1) \). It's essentially the line that just kisses the curve of the function at that spot without crossing it.