Question

Some of these can be multiplied out. For a few of these, take the derivative both before and after multiplying out, and compare the two. $$y=\sqrt{x}\left(3 x^{2}+2 x-3\right)$$

Short answer

The derivatives before and after multiplying out, \( \frac{1}{2}x^{-1/2}(3x^2 + 2x - 3) + \sqrt{x}(6x + 2)\) and \(\frac{15}{2}x^{3/2} + 3x^{1/2} - \frac{3}{2}x^{-1/2}\), result in the same expression highlighting the consistency of the derivative rules.

Step-by-step solution

Applying Product Rule for Differentiation

To take the derivative before multiplying out, apply the Product Rule, which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function, \(d(uv) = u'v + uv'\). Let \(u= \sqrt{x}\) and \(v=3x^2+2x-3\).

Differentiating Each Function Separately

Differentiate both \(u\) and \(v\) separately. For \(u=\sqrt{x}\), the derivative is \(u' = \frac{1}{2}x^{-1/2}\). For \(v=3x^2+2x-3\), the derivative is \(v' = 6x + 2\).

Applying the Product Rule

Apply the product rule using the derivatives found, \(y' = u'v + uv'\). Substituting the derivatives we get, \(y' = \frac{1}{2}x^{-1/2}(3x^2 + 2x - 3) + \sqrt{x}(6x + 2)\).

Multiplying Out the Original Function

Now, multiply out the original function, \(y = \sqrt{x}(3x^2 + 2x - 3)\), to apply the derivative to the expanded form which gives us \(y = 3x^{5/2} + 2x^{3/2} - 3x^{1/2}\).

Differentiating the Expanded Function

Take the derivative of the expanded function term by term. The derivative \(y' = \frac{5}{2} \cdot 3x^{3/2} + \frac{3}{2} \cdot 2x^{1/2} - \frac{1}{2} \cdot 3x^{-1/2}\) simplifies to \(y' = \frac{15}{2}x^{3/2} + 3x^{1/2} - \frac{3}{2}x^{-1/2}\)

Compare the Derivatives

Comparing the derivatives from applying the product rule before multiplying out and after multiplying gives the same result, confirming that the derivative rules are consistent.

Key concepts

Derivative Calculus

Derivative calculus, a fundamental aspect of calculus, is the study of how functions change when their inputs change. It's concerned with the calculation of the derivative, which gives the rate at which a function's value is changing at any given point. In essence, the derivative is the instant slope of the function at a particular point. Understanding derivatives is crucial as they appear across various fields such as physics, engineering, and economics to describe phenomena such as velocity, growth rates, and marginal costs.

To conceptualize this, imagine driving a car and looking at the speedometer, which shows your speed at that very instant—not an average over time, like the information you’d get from calculating how long a trip took. That's how the derivative displays information: it gives you the 'speed' of a function's output regarding its input value at any instant. The basic notation for a derivative is either \( f'(x) \) or \( \frac{df}{dx} \), where \( f \) is the function, and \( x \) is the variable.

Power Rule Differentiation

When you encounter a function that involves a power of \( x \), like \( x^n \), you can use the power rule to find its derivative easily. The power rule is an indispensable tool and it states that if \( f(x) = x^n \) where \( n \) is any real number, then \( f'(x) = nx^{n-1} \). This means you bring down the exponent as a coefficient in front of the variable and subtract one from the exponent to find the new power.

Example of Power Rule:

For the function \( y=x^3 \), applying the power rule would give us \( y' = 3x^{3-1} \) which simplifies to \( y' = 3x^2 \). What makes the power rule so valuable is its simplicity; it's a quick and direct method to determine the derivatives for polynomial functions. To illustrate, in our exercise, the term \( 3x^{2} \) is differentiated to \( 3 \cdot 2x^{2-1} \) or simply \( 6x \).

Chain Rule Differentiation

Sometimes functions are composed of multiple functions nested within each other, and that's where the chain rule shines. The chain rule is vital for finding the derivative of composite functions. It works on the principle that if \( u(g(x)) \) is composed of two functions, \( u \) and \( g \), the derivative \( u'(g(x)) \) is \( u' \cdot g' \). The chain rule tells us how to differentiate a function of a function, effectively 'unpacking' the nested functions layer by layer.

Application of Chain Rule:

Consider \( y=\sqrt{x}(3x^2+2x-3) \). To differentiate \(\sqrt{x}\), or \(x^{1/2}\), as a function of \(x\), we follow the power rule and get \(\frac{1}{2}x^{-1/2}\). However, the chain rule is needed to differentiate the entire expression's derivative correctly. We treat \(\sqrt{x}\) as the outer function, and \(3x^2+2x-3\) is the inner part that gets affected when the outer function changes. The chain rule helps in accounting for both these relationships in the derivative.