Problem 34
Question
In Exercises 33–36, find an equation of the tangent line to the graph of the function at the given point. $$ y=x^{\cosh x}, \quad(1,1) $$
Step-by-Step Solution
Verified Answer
The equation of the tangent line to the graph of the function at the point (1,1) is \(y = \cosh(1) \cdot x + 1 - \cosh(1)\)
1Step 1: Find the derivative of the function
The derivative of the function \(y=x^{\cosh x}\) can be found using the Chain Rule and the Product Rule. The derivative, which we'll call \(y'\), is calculated as follows: \[y' = (\cosh x) \cdot x^{(\cosh x - 1)} + x^{\cosh x} \cdot (\sinh x) \cdot \ln x\]
2Step 2: Evaluate the derivative at the given point
Evaluate the derivative at the point (1,1). When \(x=1\), \(y' = \cosh(1) + e^1 \cdot \sinh(1) \cdot \ln(1)\). Note that \(\ln(1)\) equals 0, so the second term simplifies to 0. Therefore, \(y'\) at the point (1,1) is \(\cosh(1)\).
3Step 3: Substitute the point and the derivative into the equation of a line
The equation of a line is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. To find the equation of the tangent line, substitute \(x=1\), \(y=1\), and \(m=\cosh(1)\) into this equation and solve for \(b\). Doing so gives: \(1 = \cosh(1) \cdot 1 + b\), rearranging for \(b\) gives \(b = 1 - \cosh(1)\).
Key Concepts
Chain Rule in CalculusApplying the Product RuleUnderstanding Tangent Lines
Chain Rule in Calculus
The chain rule is a fundamental concept in calculus used to find the derivative of composite functions. It allows us to differentiate functions that are composed of other functions. Imagine peeling layers like an onion; you tackle each layer one at a time.
When you see a function inside another function, that's where the chain rule comes into play. For example, if you have a function like \(f(g(x))\), the chain rule helps you discover the rate at which \(f\) changes with \(g(x)\) and how \(g(x)\) changes with \(x\).
When you see a function inside another function, that's where the chain rule comes into play. For example, if you have a function like \(f(g(x))\), the chain rule helps you discover the rate at which \(f\) changes with \(g(x)\) and how \(g(x)\) changes with \(x\).
- First, find the derivative of the outer function \(f\), treating the inner function \(g(x)\) as a constant.
- Next, multiply this by the derivative of the inner function \(g(x)\).
Applying the Product Rule
In calculus, the product rule is essential when we need to differentiate two functions that are multiplied together. This rule is like teamwork; each function takes turns to step outside for differentiation while the other remains constant.
If you have a product of functions like \(u(x) \cdot v(x)\), the product rule tells us to differentiate it as follows:
In our given function \(y=x^{\cosh x}\), we apply the product rule where one part of the function is \(x^{\cosh x}\), and the other part involves the hyperbolic cosine function \(\cosh x\). The product rule keeps everything neat and organized, simplifying even complex expressions.
If you have a product of functions like \(u(x) \cdot v(x)\), the product rule tells us to differentiate it as follows:
- Differentiating \(u(x)\) while keeping \(v(x)\) constant: \(u'(x) \cdot v(x)\).
- Then, keeping \(u(x)\) constant while differentiating \(v(x)\): \(u(x) \cdot v'(x)\).
In our given function \(y=x^{\cosh x}\), we apply the product rule where one part of the function is \(x^{\cosh x}\), and the other part involves the hyperbolic cosine function \(\cosh x\). The product rule keeps everything neat and organized, simplifying even complex expressions.
Understanding Tangent Lines
Tangent lines are critical in calculus because they give us a linear perspective of how a curve behaves at a specific point. Imagine driving on a windy mountain road. The tangent line is like going straight when at a curve point, giving you the direction of travel over a small distance.
When you find the derivative of a function at a particular point, you're actually finding the slope of the tangent line. This line touches the curve at precisely one point without crossing it (at least at that immediate vicinity).
The formula for the tangent line is \(y = mx + b\), where \(m\) is the slope found from the derivative, and \(b\) is the y-intercept. Substituting the point of interest into this formula, along with the derivative, helps give a precise equation for the tangent.
In our specific problem, calculating the derivative for \(y = x^{\cosh x}\) and then substituting the given point lets us craft its tangent line. This equation helps understand how fast and in which direction the function is changing at that point.
When you find the derivative of a function at a particular point, you're actually finding the slope of the tangent line. This line touches the curve at precisely one point without crossing it (at least at that immediate vicinity).
The formula for the tangent line is \(y = mx + b\), where \(m\) is the slope found from the derivative, and \(b\) is the y-intercept. Substituting the point of interest into this formula, along with the derivative, helps give a precise equation for the tangent.
In our specific problem, calculating the derivative for \(y = x^{\cosh x}\) and then substituting the given point lets us craft its tangent line. This equation helps understand how fast and in which direction the function is changing at that point.
Other exercises in this chapter
Problem 33
Finding an Indefinite Integral of a Trigonometric Function In Exercises \(31-40\) , find the indefinite integral. $$ \int \csc 2 x d x $$
View solution Problem 33
Show that \(f\) is strictly monotonic on the given interval and therefore has an inverse function on that interval. \(f(x)=\cos x, \quad[0, \pi]\)
View solution Problem 34
Completing the Square In Exercises \(33-42,\) find or evaluate the integral by completing the square. $$ \int_{-2}^{2} \frac{d x}{x^{2}+4 x+13} $$
View solution Problem 34
Solving an Equation In Exercises \(25-34,\) solve the equation accurate to three decimal places. $$ \log _{5} \sqrt{x-4}=3.2 $$
View solution