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Chapter 2: Problem 3

Differentiate. $$ y=x \sqrt{x}, \text { two ways } $$

Short Answer

Expert verified
The derivative is \( \frac{dy}{dx} = \frac{3}{2}\sqrt{x} \).

Step by step solution

01

Rewrite the expression

Rewrite the given function for clarity. The function given is \( y = x \sqrt{x} \). We can express this as \( y = x \times x^{1/2} \), which simplifies to \( y = x^{3/2} \).
02

Differentiate using the power rule

Using the power rule for differentiation, which states \( \frac{d}{dx}[x^n] = nx^{n-1} \), differentiate the function \( y = x^{3/2} \). Thus, \( \frac{dy}{dx} = \frac{3}{2}x^{1/2} \) or equivalently \( \frac{dy}{dx} = \frac{3}{2}\sqrt{x} \).
03

Rewrite the original function for the product rule

Rewrite the initial function in a form suitable for the product rule. The original function \( y = x \sqrt{x} \) can be written as \( y = x \cdot x^{1/2} \). Consider \( u = x \) and \( v = x^{1/2} \).
04

Differentiate using the product rule

Apply the product rule, which states \( \frac{d}{dx}[uv] = u \frac{dv}{dx} + v \frac{du}{dx} \). Find \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = \frac{1}{2}x^{-1/2} \). Substitute back to get \( \frac{dy}{dx} = x \cdot \frac{1}{2}x^{-1/2} + x^{1/2} \cdot 1 \). Simplify to get \( \frac{dy}{dx} = \frac{1}{2}x^{1/2} + x^{1/2} \), which further simplifies to \( \frac{dy}{dx} = \frac{3}{2}\sqrt{x} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule
The power rule is a fundamental tool in differentiation that simplifies the process of finding derivatives. If you have a function in the form of \(x^n\), where \(n\) is any real number, the power rule allows you to quickly find its derivative. The rule states: \[\frac{d}{dx}[x^n] = nx^{n-1} \] For example, in the function \(y = x^{3/2}\), applying the power rule involves:
  • Identifying the exponent \(n = 3/2\)
  • Multiplying the exponent by the coefficient (which is 1 if there's no visible coefficient)
  • Decreasing the exponent by 1 to get the new power of \(x\)
So, the derivative of \(x^{3/2}\) using the power rule is: \[ \frac{d}{dx}[x^{3/2}] = \frac{3}{2}x^{1/2} \] This shows how calculus makes it easier to handle complex functions and understand their behavior.
Product Rule
The product rule is essential when you are differentiating the product of two functions. Let's say you have two functions, \(u(x)\) and \(v(x)\), and you want to differentiate their product, \(u(x) \times v(x)\). The product rule states: \[ \frac{d}{dx}[uv] = u \frac{dv}{dx} + v \frac{du}{dx} \] This means you differentiate each function separately, then add the product of the first function and the derivative of the second to the product of the second function and the derivative of the first. For example, with \(y = x \times x^{1/2}\), setting \(u = x\) and \(v = x^{1/2}\), we get:
  • \

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Most popular questions from this chapter

Differentiate. $$ y=\sqrt{x+\csc x} $$

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Let \(f(x)=\frac{x^{2}}{x^{2}-1}\) and \(g(x)=\frac{1}{x^{2}-1}\). a) Compute \(f^{\prime}(x)\). b) Compute \(g^{\prime}(x)\). c) Compare your answers in parts (a) and (b) and explain.
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