Problem 28 - Calculus Early Transcendentals

Question

Compute \(dy/dx\) for the following functions. $$y=x \tanh x$$

Step-by-Step Solution

Answer

Question: Find the derivative of the function $$y = x \tanh x$$ with respect to x. Answer: The derivative of the given function is $$\frac{dy}{dx} = x\text{sech}^2 x + \tanh x$$.

Step 1: Find the derivative of u(x)

We want to find the derivative of $$u = x$$ with respect to x. This is a simple power function, and its derivative is simply 1. So, $$\frac{du}{dx} = 1$$.

Step 2: Find the derivative of v(x)

Now, we want to find the derivative of $$v = \tanh x$$ with respect to x. We know that $$\tanh x = \frac{\sinh x}{\cosh x}$$, where $$\sinh x = \frac{e^x - e^{-x}}{2}$$ and $$\cosh x = \frac{e^x + e^{-x}}{2}$$. We can use the quotient rule for derivatives, which states that for a function $$y = \frac{A(x)}{B(x)}$$, where A(x) and B(x) are differentiable functions of x, we have $$\frac{dy}{dx} = \frac{A'(x)B(x) - A(x)B'(x)}{(B(x))^2}$$. In our case, A(x) = sinh x, B(x) = cosh x. We first find the derivative of A(x) and B(x). For A'(x), we have $$\frac{d(\sinh x)}{dx} = \frac{d(\frac{e^x - e^{-x}}{2})}{dx} = \frac{e^x + e^{-x}}{2} = \cosh x$$. Similarly, for B'(x), we have $$\frac{d(\cosh x)}{dx} = \frac{d(\frac{e^x + e^{-x}}{2})}{dx} = \frac{e^x - e^{-x}}{2} = \sinh x$$. Now, using the quotient rule, we get the derivative of v(x) as: $$\frac{dv}{dx} = \frac{(\cosh x)(\cosh x) - (\sinh x)(\sinh x)}{(\cosh x)^2} = \frac{(\cosh x)^2 - (\sinh x)^2}{(\cosh x)^2} = 1 - \tanh^2 x = \text{sech}^2 x$$, where sech x is the hyperbolic secant function.

Step 3: Apply the product rule

Now that we have the derivatives of u(x) and v(x), we can apply the product rule to find the derivative of y. Recall that $$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$$. Plugging in the expressions for u, v, and their derivatives, we have: $$\frac{dy}{dx} = x\cdot\text{sech}^2 x + \tanh x\cdot1 = x\text{sech}^2 x + \tanh x$$ This is the derivative of the given function $$y=x \tanh x$$ with respect to x.

Key Concepts

Product Rule

When you have a function that is a product of two other functions, you can use the product rule to find its derivative. The product rule states that if you have two functions, say \( u(x) \) and \( v(x) \), then the derivative of their product \( y = u(x) \cdot v(x) \) is given by:

• \( \frac{dy}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} \)

This means you first take the derivative of the first function and multiply it by the second function, then add this to the first function multiplied by the derivative of the second function.
For example, if you have \( y = x \times \tanh(x) \), using the product rule involves finding the derivative of \( x \), which is \( 1 \), and the derivative of \( \tanh(x) \), which we'll discuss later. Then plug these into the formula given above.
The product rule is a foundational tool in calculus for breaking down more complex functions into simpler parts that are easier to work with and differentiate.

Quotient Rule

In calculus, the quotient rule is a method for finding the derivative of a function that is the division of two differentiable functions. If you have a function \( y = \frac{A(x)}{B(x)} \), the quotient rule tells us that:

• \( \frac{dy}{dx} = \frac{A'(x)B(x) - A(x)B'(x)}{(B(x))^2} \)

This formula helps calculate the derivative by taking the derivative of the numerator \( A(x) \) times the denominator \( B(x) \), subtracting from it the numerator \( A(x) \) times the derivative of the denominator \( B(x) \), and then dividing by the square of the denominator.
In our original problem, while calculating the derivative of \( \tanh(x) \), which can be expressed as \( \frac{\sinh(x)}{\cosh(x)} \), we apply the quotient rule. Here, \( \sinh(x) \) is \( A(x) \) and \( \cosh(x) \) is \( B(x) \).
Once applied, this yields the familiar result that the derivative of \( \tanh(x) \) is \( \text{sech}^2(x) \), another hyperbolic function.

Hyperbolic Functions

Hyperbolic functions are counterparts to the trigonometric functions, but they are based on hyperbolas instead of circles. The main hyperbolic functions include \( \sinh(x) \), \( \cosh(x) \), and \( \tanh(x) \), among others.
Here's a quick overview of the key hyperbolic functions:

• \( \sinh(x) \) or hyperbolic sine, defined as \( \frac{e^x - e^{-x}}{2} \).
• \( \cosh(x) \) or hyperbolic cosine, defined as \( \frac{e^x + e^{-x}}{2} \).
• \( \tanh(x) \) or hyperbolic tangent, given by \( \frac{\sinh(x)}{\cosh(x)} \).

Hyperbolic functions have interesting properties and identities, much like their circular cousins. When you differentiate \( \tanh(x) \), the result is \( \text{sech}^2(x) \), where \( \text{sech}(x) \) is the hyperbolic secant, defined as \( \frac{1}{\cosh(x)} \).
These functions are widely used in calculus and engineering and provide solutions to certain types of differential equations.