Question

Find \(d y / d x\). \(y=\sqrt{x^{2} \cos x}\)

Short answer

The derivative \( \frac{d y}{d x} \) is \( \frac{x \cos x - \frac{1}{2}x^2 \sin x}{\sqrt{x^2 \cos x}} \).

Step-by-step solution

Rewrite the Given Function Using Exponents

The function given is \(y = \sqrt{x^2 \cos x}\). We can rewrite this expression in terms of exponents: \(y = (x^2 \cos x)^{1/2}\). This will make it easier to use the chain rule during differentiation.

Apply the Chain Rule

The chain rule states that if we have a composition of functions, \(h(x) = f(g(x))\), then the derivative is \(h'(x) = f'(g(x)) \cdot g'(x)\). Here, let \(u = x^2 \cos x\) so that \(y = u^{1/2}\). The derivative \(d y/d x\) is \((1/2) u^{-1/2} \cdot d u/d x\).

Find the Derivative of the Inside Function

Now find the derivative of \(u = x^2 \cos x\). We will use the product rule since \(u(x)\) is the product of \(x^2\) and \(\cos x\). The product rule states that \(d(uv)/dx = u'v + uv'\). Here, let \(u = x^2\) and \(v = \cos x\). Then \(u' = 2x\) and \(v' = -\sin x\). So, \(du/dx = 2x \cdot \cos x + x^2 \cdot (-\sin x)\).

Combine the Results

Substitute \(du/dx\) back into the chain rule expression from Step 2. We have \(dy/dx = (1/2) (x^2 \cos x)^{-1/2} \cdot (2x \cos x - x^2 \sin x)\). Simplify this expression to get the final derivative.

Simplify the Derivative Expression

Simplifying further, \(dy/dx = \frac{1}{2} (x^2 \cos x)^{-1/2} \cdot (2x \cos x - x^2 \sin x)\) can be written in a simplified form: \(dy/dx = \frac{x \cos x - \frac{1}{2}x^2 \sin x}{\sqrt{x^2 \cos x}}\).

Key concepts

Understanding the Chain Rule

The chain rule is a fundamental concept in calculus used when differentiating composite functions. A composite function is a function inside another function, like a nested "mathematical doll." In simple terms, when you have a function of the form \( h(x) = f(g(x)) \), the chain rule helps us find its derivative.
To apply the chain rule, you first differentiate the outer function \( f \) with respect to the inner function \( g(x) \), and then multiply by the derivative of the inner function. Mathematically, this can be expressed as:

• \( h'(x) = f'(g(x)) \cdot g'(x) \)

In the problem provided, the chain rule is applied to \( y = (x^2 \cos x)^{1/2} \). By letting \( u = x^2 \cos x \), we simplify the problem as \( y = u^{1/2} \). The chain rule then tells us to first find the derivative of \( u^{1/2} \) with respect to \( u \), which is \( (1/2)u^{-1/2} \), and then multiply it by \( du/dx \).
This step-by-step approach makes complex derivatives more manageable.

Applying the Product Rule

The product rule is another crucial tool in our differentiation toolkit. It is used when finding the derivative of the product of two functions. If you have a product \( y = u(x)v(x) \), the product rule states:

• \( \frac{d}{dx}[uv] = u'v + uv' \)

This rule dictates that you take the derivative of the first function \( u(x) \), multiply it by the second function \( v(x) \), then add the product of the first function and the derivative of the second function.
In the exercise, we are tasked with differentiating \( u = x^2 \cos x \). Applying the product rule here involves letting \( u = x^2 \) and \( v = \cos x \), finding their derivatives: \( u' = 2x \) and \( v' = -\sin x \). So, the derivative \( du/dx = 2x \cdot \cos x + x^2 \cdot (-\sin x) \).
This step is essential because it provides the \( du/dx \) term needed to complete the differentiation process initiated by the chain rule.

Derivative Simplification: Bringing It All Together

Once the various rules and techniques have been applied, the final step in differentiation is often simplification. Simplifying derivatives allows us to express the result in the most compact and elegant form possible, making it more understandable and useful.
In the original exercise, after applying the chain and product rules, the resulting expression \( \frac{dy}{dx} = \frac{1}{2} (x^2 \cos x)^{-1/2} \cdot (2x \cos x - x^2 \sin x) \) might still seem a bit complex. Therefore, our goal is to rewrite it in a more simplified and clear form.

• The expression is simplified to: \( \frac{x \cos x - \frac{1}{2}x^2 \sin x}{\sqrt{x^2 \cos x}} \).

This simplification makes it clear how the various pieces of the derivative interact with each other. Through simplification, you achieve a more intuitive understanding of how changes in \( x \) affect \( y \).
It is these steps that transform a potentially messy-looking derivative into a succinct and insightful expression.