Problem 17 - Calculus Single Variable

Question

An expression for \(f(x)\) is given. Compute the first, second, and third derivatives of \(f(x)\) with respect to \(x\). \(27\left(x^{2}+1\right)^{5 / 3}\)

Step-by-Step Solution

Answer

The derivatives are: 1st: \( f'(x) = 90x(x^2 + 1)^{2/3} \) 2nd: \( f''(x) = 90(x^2+1)^{2/3} + 120x^2(x^2+1)^{-1/3} \) 3rd: \( f'''(x) = 60x(x^2+1)^{-1/3} + 180(x^2+1)^{2/3} - 40x^3(x^2+1)^{-4/3} \)."

Step 1: Understand the Given Function

We are given the function \( f(x) = 27(x^{2}+1)^{5/3} \). Our task is to compute the first, second, and third derivatives of this function with respect to \( x \). We first notice that this function is a composition of functions and will require the use of the chain rule for differentiation.

Step 2: Differentiate to Find the First Derivative

To find the first derivative \( f'(x) \), we apply the chain rule. Let \( u = x^2 + 1 \), then \( f(x) = 27u^{5/3} \). The derivative of \( 27u^{5/3} \) with respect to \( u \) is \( 45u^{2/3} \). Also, \( \frac{du}{dx} = 2x \). By the chain rule, \( f'(x) = \frac{d}{dx}[27u^{5/3}] = 45u^{2/3} \cdot 2x = 90x(x^2 + 1)^{2/3} \).

Step 3: Differentiate to Find the Second Derivative

We differentiate \( f'(x) = 90x(x^2 + 1)^{2/3} \). Using the product rule and chain rule: \[ f''(x) = 90\left[ (x^2 + 1)^{2/3} + x \cdot \frac{2}{3}(x^2 + 1)^{-1/3} \cdot 2x \right] \]\[ = 90(x^2+1)^{2/3} + 120x^2(x^2+1)^{-1/3} \].

Step 4: Differentiate to Find the Third Derivative

Now for the third derivative: differentiate \( f''(x) = 90(x^2+1)^{2/3} + 120x^2(x^2+1)^{-1/3} \). This again requires the product and chain rules applied to each term:\[ f'''(x) = 60x(x^2+1)^{-1/3} + 180(x^2+1)^{2/3} - 40x^3(x^2+1)^{-4/3} \].

Key Concepts

Chain Rule in Differentiation

Differentiating functions that are compositions of two or more functions often require the chain rule. Imagine you have a function nested inside another function, much like a matryoshka doll. For example, when you have functions of the form \( (x^2 + 1)^{5/3} \), the chain rule helps separate out each component to make differentiation manageable.
Here's how the chain rule works:

• Identify the two functions: an "outer function" and an "inner function".
• Differentiation involves differentiating the outer function with respect to the inner function, and then multiplying by the derivative of the inner function.

In our exercise, the inner function \( u \) is \( x^2 + 1 \), and the outer function is \( 27u^{5/3} \). This rule helps by breaking down complex functions into simpler parts. The chain rule reveals how changes in one part impact the entire function, ensuring smoother computation and better understanding.

Product Rule in Calculus

When calculations involve multiplying two functions, the product rule is your best friend. For example, when you differentiated \( 90x(x^2 + 1)^{2/3} \), you employed this essential rule. The product rule is essential because differentiation directly applied to a product of functions doesn't yield the correct derivative.
The product rule works like this:

• Given two functions, say \( u(x) \) and \( v(x) \).
• The derivative of their product \( u(x) \, v(x) \) is \( u'(x) \, v(x) + u(x) \, v'(x) \).

By applying it to our function's differentiation, you ensure no parts are left out from your calculations. The product rule is vital when dealing with complex, multi-part functions, just like the combination of \( 90x \) and \( (x^2 + 1)^{2/3} \). Whenever you see two parts multiplied, switch on your product rule thinking.

Understanding Higher-Order Derivatives

Higher-order derivatives allow us to dive deeper into the behavior of functions. In the context of this exercise, you were asked not just to find the first derivative, but also the second and third derivatives. These derivatives speak volumes about the function's curvature and inflection.
Here is what they signify:

• The first derivative \( f'(x) \) provides the rate of change or slope of the function. It tells you how fast a function is changing at a given point.
• The second derivative \( f''(x) \) informs about the concavity of the function. It's used to determine if a graph is curving upwards or downwards.
• The third derivative \( f'''(x) \) can offer insights into the function's jerk or the rate of change of the acceleration. Although it’s less frequently required, it becomes crucial in certain analyses.

Through applying the principles of calculus, higher-order derivatives are crucial tools in understanding the deeper nature of mathematical models, providing insights that are beyond the reach of basic computation.