Question

Compute \(d y / d x\) for the following functions. \(y=\tanh ^{2} x\)

Short answer

Question: Find the derivative of the function \(y=\tanh^2(x)\) with respect to \(x\). Answer: The derivative of the function \(y=\tanh^2(x)\) with respect to \(x\) is \(\frac{dy}{dx} = 2\tanh(x) \cdot \frac{1}{\cosh^2(x)}\).

Step-by-step solution

Identify the outer and inner functions

The given function is \(y=\tanh^2(x)\). The outer function is \(f(u)=u^2\) and the inner function is \(g(x)=\tanh(x)\). Our goal is to find the derivative of \(y\) with respect to \(x\).

Apply the chain rule

According to the chain rule: \[ \frac{dy}{dx} = \frac{d(f \circ g)}{dx} = \frac{df}{du}\cdot\frac{dg}{dx} \]

Find the derivative of the outer function

To find the derivative of the outer function \(f(u)=u^2\), we differentiate it with respect to \(u\): \[ \frac{df}{du} = \frac{d(u^2)}{du}=2u\]

Find the derivative of the inner function

To find the derivative of the inner function \(g(x)=\tanh(x)\), we differentiate it with respect to \(x\): \[ \frac{dg}{dx} = \frac{d(\tanh(x))}{dx}= \frac{1}{\cosh^2(x)}\]

Substitute the derivatives in the chain rule formula

After finding the derivatives of the outer and inner functions, we substitute them into the chain rule formula: \[ \frac{dy}{dx} = \frac{df}{du}\cdot\frac{dg}{dx} = 2u \cdot \frac{1}{\cosh^2(x)}\]

Replace \(u\) with the inner function

Now, replace \(u\) with the inner function \(g(x)=\tanh(x)\): \[ \frac{dy}{dx} = \frac{1}{\cosh^2(x)} \cdot 2 \cdot \tanh(x)\]

Write the final answer

So the derivative of \(y=\tanh^2(x)\) with respect to \(x\) is: \[ \frac{dy}{dx} = 2\tanh(x) \cdot \frac{1}{\cosh^2(x)}\]

Key concepts

Chain Rule

The chain rule is a fundamental technique in calculus for finding the derivative of composite functions. Understanding how it works can greatly simplify complex differentiation tasks.

• The chain rule states that if you have a function that is composed of two functions, say \( f \) and \( g \), then the derivative of the composite function is found by multiplying the derivative of \( f \) with respect to its argument, by the derivative of \( g \) with respect to its argument.
• For example, if \( y = (f \circ g)(x) \), then the derivative \( \frac{dy}{dx} \) is given by \( \frac{df}{du} \cdot \frac{dg}{dx} \).
• This approach is particularly useful when dealing with functions that are nested or when one function forms the input of another.

In the given problem, the function \( y = \tanh^2(x) \) is a composition. Here, the outer function is \( u^2 \), and the inner function is \( \tanh(x) \). The task involves differentiating both functions and then applying the chain rule.
By applying this rule, we multiply the derivative of the outer function evaluated at the inner function by the derivative of the inner function itself.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola rather than a circle. These functions are essential in various fields such as calculus, complex analysis, and certain areas of engineering and physics.

• The hyperbolic functions include \( \sinh \), \( \cosh \), \( \tanh \), and their reciprocals.
• Each has its own set of properties that closely mimic trigonometric functions, such as the identity \( \cosh^2(x) - \sinh^2(x) = 1 \).
• In our derivative problem, we deal with \( \tanh(x) \), the hyperbolic tangent, whose derivative is \( \frac{1}{\cosh^2(x)} \).

When differentiating hyperbolic functions, it is often helpful to recall these derivatives directly as they parallel the properties of trigonometric derivatives.
This knowledge aids in problems like the given one, where recognizing and applying the derivative of \( \tanh(x) \) is crucial.

Differentiation Techniques

Differentiation involves finding the rate at which a function is changing at any point. It is a key concept in calculus, providing the tools for analyzing changes.

• For straightforward functions, direct rules such as the power rule or product rule can be used.
• However, for more complex functions, like compositions, the chain rule plays a pivotal role. It allows us to break down functions into manageable parts, differentiating each part first and then combining them.
• The exercise demonstrates the application of these techniques by first identifying the inner and outer functions and then differentiating each individually.

Mastering these techniques is fundamental, as they are used extensively in both academic problems and real-world applications.
Overall, learning to skillfully apply chain rule and understand specific functions such as hyperbolic ones can turn seemingly daunting problems into manageable tasks.