Question

Find \(D_{x} y\). $$ y=x \cosh ^{-1}(3 x) $$

Short answer

\( D_x y = \cosh^{-1}(3x) + \frac{3x}{\sqrt{9x^2 - 1}} \).

Step-by-step solution

Identify Functions Involved

The expression for the function is given as \( y = x \cosh^{-1}(3x) \). This expression is a product of two functions. The first function is \( u(x) = x \) and the second function is \( v(x) = \cosh^{-1}(3x) \). We'll use the product rule to find \( D_x y \).

Apply the Product Rule

The product rule for differentiation states that if \( y = u \cdot v \), then \( D_x y = u'v + uv' \). Apply this rule to find the derivative of \( y = x \cosh^{-1}(3x) \).

Differentiate the First Function

First function: \( u(x) = x \). The derivative of \( u \) with respect to \( x \) is \( u'(x) = 1 \).

Differentiate the Second Function

Second function: \( v(x) = \cosh^{-1}(3x) \). The derivative of \( \cosh^{-1}(x) \) is \( \frac{d}{dx} [\cosh^{-1}(x)] = \frac{1}{\sqrt{x^2 - 1}} \). Use the chain rule to differentiate \( \cosh^{-1}(3x) \):\[ v'(x) = \frac{d}{dx} \left[ \cosh^{-1}(3x) \right] = \frac{1}{\sqrt{(3x)^2 - 1}} \cdot 3 = \frac{3}{\sqrt{9x^2 - 1}} \].

Substitute into Product Rule

Substitute \( u'(x) = 1 \), \( v(x) = \cosh^{-1}(3x) \), and \( v'(x) = \frac{3}{\sqrt{9x^2 - 1}} \) into the product rule formula:\[ D_x y = 1 \cdot \cosh^{-1}(3x) + x \cdot \frac{3}{\sqrt{9x^2 - 1}} \].

Simplify the Expression

Combine and simplify the terms to get:\[ D_x y = \cosh^{-1}(3x) + \frac{3x}{\sqrt{9x^2 - 1}} \]. This is the derivative of the given function.

Key concepts

Product Rule

When you are trying to differentiate a product of two functions, the product rule is your best friend. It's like a tool that helps you handle two snapshots at once. If a function is a product of two smaller functions, say \( y = u(x) \cdot v(x) \), then the derivative of \( y \), written as \( D_x y \), is calculated as \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \).

In our problem, we see the expression \( y = x \cdot \cosh^{-1}(3x) \). You can think of \( x \) as one piece and \( \cosh^{-1}(3x) \) as another. Apply the product rule here by identifying:

• First function: \( u(x) = x \)
• Second function: \( v(x) = \cosh^{-1}(3x) \)

Understanding this makes it easier to compute derivatives without getting mixed up. You just need to figure out the derivatives of both functions involved and then plug them into the product rule formula.

Hyperbolic Functions

Hyperbolic functions are similar to trigonometric functions but deal with hyperbolas instead of circles. The function \( \cosh^{-1}(x) \), called the inverse hyperbolic cosine, mirrors certain properties of the inverse cosine but for hyperbolas.

Unlike trigonometric functions, hyperbolic functions often relate to exponential growth and decay phenomena. Specifically, the formula for the derivative of \( \cosh^{-1}(x) \) is \( \frac{1}{\sqrt{x^2 - 1}} \). This expression tells you how the inverse hyperbolic cosine changes at a given point and is crucial in differentiation involving these functions.

In the exercise, you have \( \cosh^{-1}(3x) \) and to find its derivative, it's important to apply the chain rule, which allows you to easily handle the intricacies of the outer and inner functions combined.

Chain Rule

The chain rule is a differentiation technique used for composite functions—functions nested within each other like layers of an onion. It’s very handy when you need to deal with complex expressions involving functions applied one inside the other, like \( \cosh^{-1}(3x) \).

The chain rule states that to differentiate a composite function \( f(g(x)) \), you multiply the derivative of the outer function evaluated at the inner function by the derivative of the inner function. In formula terms, if \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

For \( v(x) = \cosh^{-1}(3x) \), let's take it apart:

• Outer function: \( \cosh^{-1}(x) \)
• Inner function: \( 3x \)

First, differentiate the outer function \( \cosh^{-1} \), then multiply by the derivative of the inner function which is 3. You end up with \( \frac{3}{\sqrt{9x^2 - 1}} \) as the derivative of our second function. Using the chain rule turns complex differentiation into a manageable task and is essential for solving problems like these effectively.