Problem 33 - Calculus and its applications

Question

Differentiate each function. \(f(x)=6 x^{-4}\left(6 x^{3}+10 x^{2}-8 x+3\right)\)

Step-by-Step Solution

Answer

The derivative is \( f'(x) = -36x^{-2} - 120x^{-3} + 144x^{-4} - 72x^{-5} \).

Step 1: Use the Product Rule

The function is the product of two expressions: \(u(x) = 6x^{-4}\) and \(v(x) = 6x^3 + 10x^2 - 8x + 3\). We'll use the product rule, which states that if \(f(x) = u(x)v(x)\), then \(f'(x) = u'(x)v(x) + u(x)v'(x)\). We need to find the derivatives \(u'(x)\) and \(v'(x)\).

Step 2: Differentiate \(u(x)\)

To find \(u'(x)\), differentiate \(6x^{-4}\) using the power rule. The power rule states \(\frac{d}{dx}[x^n] = n x^{n-1}\). Thus, \(u'(x) = 6(-4)x^{-5} = -24x^{-5}\).

Step 3: Differentiate \(v(x)\)

The function \(v(x) = 6x^3 + 10x^2 - 8x + 3\) is a polynomial. Differentiate term by term:- \(\frac{d}{dx}[6x^3] = 18x^2\)- \(\frac{d}{dx}[10x^2] = 20x\)- \(\frac{d}{dx}[-8x] = -8\)- \(\frac{d}{dx}[3] = 0\)Combining these, \(v'(x) = 18x^2 + 20x - 8\).

Step 4: Apply the Product Rule

Substitute \(u'(x)\), \(v(x)\), \(u(x)\), and \(v'(x)\) into the product rule formula:\[ f'(x) = (-24x^{-5})(6x^3 + 10x^2 - 8x + 3) + (6x^{-4})(18x^2 + 20x - 8)\]

Step 5: Simplify the Expression

Expand and simplify both terms:- The first term: \((-24x^{-5})(6x^3 + 10x^2 - 8x + 3)\) becomes: - \(-24 \cdot 6x^{-2} = -144x^{-2}\) - \(-24 \cdot 10x^{-3} = -240x^{-3}\) - \(-24 \cdot (-8)x^{-4} = 192x^{-4}\) - \(-24 \cdot 3x^{-5} = -72x^{-5}\)- The second term: \((6x^{-4})(18x^2 + 20x - 8)\) becomes: - \(6 \cdot 18x^{-2} = 108x^{-2}\) - \(6 \cdot 20x^{-3} = 120x^{-3}\) - \(6 \cdot (-8)x^{-4} = -48x^{-4}\)Combine all terms to obtain:\[ f'(x) = (-144x^{-2} - 240x^{-3} + 192x^{-4} - 72x^{-5}) + (108x^{-2} + 120x^{-3} - 48x^{-4})\]

Step 6: Combine Like Terms

Finally, combine like terms:- \((-144x^{-2} + 108x^{-2}) = -36x^{-2}\)- \((-240x^{-3} + 120x^{-3}) = -120x^{-3}\)- \((192x^{-4} - 48x^{-4}) = 144x^{-4}\)- The term \(-72x^{-5}\) remains unchanged.Thus, the derivative is:\[ f'(x) = -36x^{-2} - 120x^{-3} + 144x^{-4} - 72x^{-5} \]

Key Concepts

Product Rule

Differentiating functions can often involve the use of the product rule. This rule applies when you're dealing with the product of two functions, say \(u(x)\) and \(v(x)\). The product rule states that the derivative of a product \(f(x) = u(x) v(x)\) is given by \(f'(x) = u'(x)v(x) + u(x)v'(x)\). Effectively, you differentiate each function separately, then sum two products: one where the first function is differentiated and the second is not, and another where the second function is differentiated and the first is not.

When solving our original exercise, the product rule is crucial. We identify \(u(x) = 6x^{-4}\) and \(v(x) = 6x^3 + 10x^2 - 8x + 3\). After differentiating both \(u(x)\) and \(v(x)\), we insert these into the product rule formula. It ensures a comprehensive derivative without missing any interaction between these parts. So, always remember: if you see a multiplication between two functions, think product rule.

Power Rule

The power rule is an essential tool for differentiation, especially when dealing with terms where the variable is raised to a power. The power rule states that for a function \(x^n\), the derivative is \(nx^{n-1}\).

This is particularly straightforward and extremely useful. In the exercise, we applied the power rule to find \(u'(x)\). For \(6x^{-4}\), using the power rule gives us \(-24x^{-5}\) because multiplying the coefficient \(6\) with the exponent \(-4\) leads to \(-24\), and you decrease the exponent by one.

The same rule helps when differentiating a polynomial, such as \(v(x)\). Each term in the polynomial, like \(6x^3\), is differentiated term by term using the power rule. For example, \(6x^3\) becomes \(18x^2\). The systematic application of the power rule allows us to deconstruct any polynomial efficiently.

Polynomial Differentiation

Polynomial differentiation builds on the power rule, making it a pillar of calculus. With a polynomial, such as \(v(x) = 6x^3 + 10x^2 - 8x + 3\), differentiation involves applying the power rule to each individual term:

• The term \(6x^3\) becomes \(18x^2\).
• The term \(10x^2\) becomes \(20x\).
• The linear term \(-8x\) becomes \(-8\).
• The constant \(3\) has no variable and thus its derivative is \(0\).

In essence, polynomial differentiation is the process of adjusting each term according to the power rule. By analyzing the polynomial term by term, you ensure that the derivative accounts for each part of the original expression. Through these methods, deriving any polynomial becomes a routine task, even for intricate expressions.