Problem 10 - Calculus of a Single Variable

Question

Finding a Derivative In Exercises \(7-34,\) find the derivative of the function. $$ f(t)=(9 t+2)^{2 / 3} $$

Step-by-Step Solution

Answer

The derivative of the function \(f(t)=(9 t+2)^{2 / 3}\) with respect to \(t\) is \(f'(t)=6(9t+2)^{-1 / 3}\).

Step 1: Identify the Inner and Outer Functions

We can write the function \(f(t)=(9t+2)^{2 / 3}\) as a composition of two functions. The inner function is \(g(t)=9t+2\) and the outer function can be written as \(h(u)=u^{2/3}\) where \(u=g(t)\).

Step 2: Find the Derivative of the Inner Function

Taking derivative of \(g(t)=9t+2\) with respect to \(t\) gives \(g'(t)=9\).

Step 3: Find the Derivative of the Outer Function

Taking derivative of \(h(u)=u^{2 / 3}\) with respect to \(u\) gives \(h'(u)=(2/3)u^{-1/3}\).

Step 4: Apply the Chain Rule

According to the chain rule, the derivative of a composite function is equal to the derivative of the outer function evaluated at the inner function times the derivative of the inner function. This gives \(f'(t)=h'(g(t)) \cdot g'(t)\). Substituting the derived values in gives us \(f'(t)=(2/3)(9t+2)^{-1 / 3} \cdot 9\).

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used to find the derivative of composite functions. When a function is the result of another function nested inside it, that's when you need the chain rule to differentiate. Imagine you have a nesting doll, with multiple layers hidden inside. Differentiating such a composed function is akin to unraveling these nested layers one by one.

To use the chain rule, identify the inner function and the outer function. Let's say the inner function is called 'g(t)' and the outer function is 'h(g(t))'. The derivative of this composite function with respect to 't' is found by taking the derivative of the outer function with respect to the inner function—denoted as 'h'(g(t))'—and multiplying it by the derivative of the inner function with respect to 't', denoted as 'g'(t)'. In short, the formula is written as \( \frac{dh}{dt} = \frac{dh}{dg} \cdot \frac{dg}{dt} \) where \( \frac{dh}{dg} \) is the derivative of 'h' with respect to 'g', and \( \frac{dg}{dt} \) is the derivative of 'g' with respect to 't'.

This rule is essential when dealing with functions within functions, allowing us to 'unpack' the derivative one layer at a time.

Derivative of Composite Functions

Understanding the derivative of composite functions is crucial when dealing with more complex calculus problems. A composite function can be thought of like a two-step process: you apply one function, and then you apply another function to the result of the first. For example, if you first put on socks (function 'g') and then put on shoes (function 'h'), your final outfit (composite function 'h∘g') depends on both actions, but in a specific order.

In mathematical terms, if you have two functions, 'g(t)' and 'h(u)', where 'u=g(t)', then the composite function 'f(t)=h(g(t))' represents this two-step process. To find the derivative of 'f' with respect to 't', we differentiate each function separately and then use the chain rule to combine these derivatives. The step-by-step process highlighted in the original exercise is a clear example of how to break down a composite function and find its derivative methodically.

Power Rule for Derivatives

The power rule is a quick and handy tool for finding the derivative of functions that are powers of x. Stated simply, if you have a function in the form \( f(x) = x^n \), where 'n' is any real number, then the derivative of the function is \( f'(x) = nx^{n-1} \). This rule tells us to bring down the exponent as a coefficient in front of the variable, and then subtract one from the exponent.

For example, in our exercise, the outer function \( h(u)=u^{2/3} \) is a candidate for the power rule. Applying it, we find the derivative \( h'(u)=(2/3)u^{-1/3} \). It's crucial to notice that this rule applies to any real-numbered exponent, not just whole numbers, allowing it to be a versatile and powerful tool in finding derivatives. However, remember that for it to be effective, the base of the power needs to be the variable of differentiation; otherwise, the chain rule must be employed to handle more complex scenarios.