Question

Compute \(d y / d x\) for the following functions. \(y=x \tanh x\)

Short answer

Answer: The derivative of the function \(y = x \tanh x\) with respect to \(x\) is \(\frac{dy}{dx} = \tanh x + x \sech^2 x\).

Step-by-step solution

Recall the Product Rule

The Product Rule states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. In mathematical notation, for two functions u(x) and v(x), the derivative of their product is: $$(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)$$

Identify the Functions u(x) and v(x)

In our given function, \(y = x \tanh x\), we have: $$u(x) = x$$ $$v(x) = \tanh x$$

Find the Derivatives u'(x) and v'(x)

Now, we need to find the derivatives of u(x) and v(x) with respect to \(x\). For u(x) = x, the derivative is simply: $$u'(x) = 1$$ For v(x) = \(\tanh x\), we need to recall the derivative of the hyperbolic tangent function, which is: $$v'(x) = \sech^2 x$$

Apply the Product Rule

Now we can apply the Product Rule to find the derivative of our given function, \(y = x \tanh x\). First, plug in the derivatives we found, and the original functions u(x) and v(x), into the Product Rule formula: $$(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)$$ $$\frac{dy}{dx} = 1 \cdot \tanh x + x \cdot \sech^2 x$$

Simplify the Expression

The derivative of the given function can be written as: $$\frac{dy}{dx} = \tanh x + x \sech^2 x$$ This is the final result for the derivative of the function \(y = x \tanh x\).

Key concepts

Product Rule

The product rule is a fundamental concept in calculus used to find the derivative of a product of two functions. This rule is quite helpful because it allows us to handle more complex functions than we could otherwise. The basic idea of the product rule can be understood as follows: if you have two differentiable functions, say \( u(x) \) and \( v(x) \), the derivative of their product \( y = u(x) \cdot v(x) \) is equal to the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second function. In a formula, it looks like this:

• \( (u(x)v(x))' = u'(x)v(x) + u(x)v'(x) \)

This is key when you are working with combinations of functions, as it gives a straightforward way to take derivatives without having to expand or simplify excessively before differentiating.

Hyperbolic Functions

Hyperbolic functions, such as hyperbolic tangent \( \tanh x \), are very similar to the more familiar trigonometric functions, but they are based on the hyperbola rather than the circle. They appear in many areas of calculus and have properties that resemble those of sine and cosine functions. What's interesting about hyperbolic functions is their connection to exponential functions. For example, the hyperbolic tangent function is given by:

• \( \tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}} \)

When it comes to differentiation, hyperbolic functions have their own set of rules. For instance, the derivative of \( \tanh x \) is \( \sech^2 x \), where \( \sech x \) is the hyperbolic secant. This can be quite useful and allows us to simplify derivatives involving hyperbolic functions in calculus problems.

Calculus

Calculus is the mathematical study of change, and it provides the tools for understanding and modeling continuous change. The two main branches of calculus are differential calculus and integral calculus.

• Differential calculus deals with the concept of a derivative, which measures how a function changes as its input changes. The derivative is a fundamental tool for finding rates of change and understanding the behavior of functions.
• Integral calculus involves the accumulation of quantities, such as areas under a curve or total accumulated change. It is essentially the reverse process of differentiation.

In the context of finding the derivative of the function \( y = x \tanh x \), calculus and the application of rules such as the product rule and the properties of hyperbolic functions allow us to understand and quantify the changes in \( y \) as \( x \) changes. This insight has numerous practical applications, from physics to engineering and beyond, where understanding change is crucial.