Problem 28
Question
Find \(d y / d x\). \(y=\sqrt{x^{2} \cos x}\)
Step-by-Step Solution
Verified Answer
The derivative \( \frac{d y}{d x} \) is \( \frac{x \cos x - \frac{1}{2}x^2 \sin x}{\sqrt{x^2 \cos x}} \).
1Step 1: 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.
2Step 2: 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 \((1/2) u^{-1/2} \cdot d u/d x\).
3Step 3: 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, \(du/dx = 2x \cdot \cos x + x^2 \cdot (-\sin x)\).
4Step 4: 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.
5Step 5: 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 RuleApplying the Product RuleDerivative Simplification: Bringing It All Together
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:
This step-by-step approach makes complex derivatives more manageable.
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:
- \( h'(x) = f'(g(x)) \cdot g'(x) \)
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:
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.
- \( \frac{d}{dx}[uv] = u'v + uv' \)
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.
It is these steps that transform a potentially messy-looking derivative into a succinct and insightful expression.
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: \( \frac{x \cos x - \frac{1}{2}x^2 \sin x}{\sqrt{x^2 \cos x}} \).
It is these steps that transform a potentially messy-looking derivative into a succinct and insightful expression.
Other exercises in this chapter
Problem 28
Find \(D_{x} y\). $$ y=e^{x} \arcsin x^{2} $$
View solution Problem 28
$$ \underline{\phantom{xxx}} , \text { find the indicated derivative. } $$ $$ D_{x} e^{2 x^{2}-x} $$
View solution Problem 28
An object is moving along a horizontal coordinate line according to the formula \(s=f(t)\), where \(s\), the directed distance from the origin, is in feet and \
View solution Problem 28
Find the indicated derivative. $$ \frac{d y}{d t}, \text { where } y=\left[\sin t \tan \left(t^{2}+1\right)\right] $$
View solution