Problem 27 - Thomas Calculus

Question

Find the derivatives of the functions in Exercises \(19-38\) $$ y=\frac{1}{21}(3 x-2)^{7}+\left(4-\frac{1}{2 x^{2}}\right)^{-1} $$

Step-by-Step Solution

Answer

The derivative is \( (3x-2)^6 - x^{-3}(4-\frac{1}{2}x^{-2})^{-2} \).

Step 1: Differentiate the First Term

Let's start by differentiating the first term of the function: \( y_1 = \frac{1}{21}(3x-2)^{7} \). We'll use the chain rule and power rule for this part. Differentiating, we get: \(\begin{align*}y_1' &= \frac{1}{21} \cdot 7(3x-2)^{6} \cdot (3),\end{align*}\) which simplifies to \( y_1' = \frac{7}{21} \cdot 3(3x-2)^{6} = \frac{1}{1}(3x-2)^{6} = (3x-2)^{6}. \)

Step 2: Differentiate the Second Term

Now, let's differentiate the second term \( y_2 = \left(4-\frac{1}{2x^{2}}\right)^{-1} \). We will use the chain rule and first rewrite the expression as \( y_2 = \left(4-\frac{1}{2}x^{-2}\right)^{-1} \). Applying the chain rule:\(\begin{align*} y_2' &= -1\left(4-\frac{1}{2}x^{-2}\right)^{-2} \cdot \frac{d}{dx}\left(4-\frac{1}{2}x^{-2}\right),\end{align*}\)Next, we differentiate the inside function:\(\begin{align*}\frac{d}{dx}\left(4-\frac{1}{2}x^{-2}\right) &= 0 + \frac{1}{2} \cdot 2x^{-3} = x^{-3}.\end{align*}\) So, \( y_2' = -x^{-3}\left(4-\frac{1}{2}x^{-2}\right)^{-2}. \)

Step 3: Combine the Derivatives

Combine the derivatives from Steps 1 and 2. Thus, the derivative of the main function \( y \) becomes: \(\begin{align*}y' &= y_1' + y_2' \&= (3x-2)^{6} - x^{-3}\left(4-\frac{1}{2}x^{-2}\right)^{-2}.\end{align*} \) This gives us the complete derivative of the function.

Key Concepts

Chain Rule

The Chain Rule is a fundamental technique in calculus for finding the derivative of a composite function. Imagine you have a function nested inside another function. The Chain Rule helps you "unravel" this by focusing on how each part changes relative to the other.

• Think of the composite function as \( f(g(x)) \), where you need to differentiate each part separately.
• The rule states: \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).

In this context, the derivative of the outer function \( f \) is multiplied by the derivative of the inner function \( g \). The Chain Rule basically says: first differentiate the outside, keep the inside the same, and then multiply by the derivative of the inside.

In our exercise, the Chain Rule was used twice. First in differentiating \( (3x-2)^7 \) and then while handling \( \left(4-\frac{1}{2x^{2}}\right)^{-1} \). In each case, the function was considered as an outer function and the term inside the brackets as the inner function.

Power Rule

The Power Rule is one of the simplest and most widely used rules in differentiation. It makes finding the derivative of a power function straightforward. If you have a term of the form \( x^n \), the Power Rule states:

• The derivative is \( nx^{n-1} \).
• Simply bring down the exponent as a coefficient and decrease the exponent by one.

In our problem, the Power Rule was used when differentiating \( (3x-2)^7 \). After applying the Chain Rule, the exponent 7 was brought down as a coefficient, and then the power was decreased by one to give us \( 6 \). This led to \( 7(3x-2)^6 \) before simplifying to \( (3x-2)^6 \).

The Power Rule is a quick tool that becomes even more powerful when combined with other rules like the Chain Rule, helping us simplify what might initially seem complex.

Derivative of a Function

A derivative captures the rate at which a function is changing at any point. It's the backbone of calculus, letting us analyze not just how things change, but the manner and speed of these changes.

• The derivative \( f'(x) \) of a function \( f(x) \) gives the slope of the tangent at any point on its curve.
• It helps answer questions about motion, growth, and optimization.

In the exercise, we derived the function \( y \) by breaking it into two manageable pieces: \( y_1 = \frac{1}{21}(3x-2)^7 \) and \( y_2 = \left(4-\frac{1}{2x^{2}}\right)^{-1} \). Each part was differentiated using the appropriate rules. Finally, the complete derivative was simply the sum \( y' = y_1' + y_2' \).

This exploration shows how derivatives allow for an in-depth look at functions from everyday math to complex applications, providing tools to predict and understand behavior in various scientific and engineering contexts.