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Find the derivatives of the functions in Exercises \(19-40\) $$q=\sqrt[3]{2 r-r^{2}}$$

Short Answer

Expert verified
The derivative is \( \frac{2(1-r)}{3(2r - r^2)^{2/3}} \).

Step by step solution

01

Rewrite the Function

First, rewrite the function in a form that makes differentiation easier. The given function is \( q = \sqrt[3]{2r - r^2} \). Rewrite this as \( q = (2r - r^2)^{1/3} \). This form makes it clear that you will need to use the chain rule to differentiate it.
02

Apply the Chain Rule

To find \( \frac{dq}{dr} \), apply the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. Here, the outer function is \( u^{1/3} \) and the inner function is \( u = 2r - r^2 \).
03

Differentiate the Outer Function

Differentiate the outer function \( u^{1/3} \) with respect to \( u \), which gives \( \frac{d}{du} u^{1/3} = \frac{1}{3} u^{-2/3} \).
04

Differentiate the Inner Function

Next, differentiate the inner function \( u = 2r - r^2 \) with respect to \( r \). This gives \( \frac{du}{dr} = 2 - 2r \).
05

Apply the Chain Rule

Combine the results from Steps 3 and 4 using the chain rule. Multiply \( \frac{1}{3} (2r - r^2)^{-2/3} \) by \( 2 - 2r \). This gives the derivative \[ \frac{dq}{dr} = \frac{1}{3}(2r - r^2)^{-2/3} (2 - 2r) \].
06

Simplify the Expression

Simplify the expression for the derivative. Since \( 2 - 2r = 2(1-r) \), the derivative simplifies to \[ \frac{dq}{dr} = \frac{2(1-r)}{3(2r - r^2)^{2/3}} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule is an essential tool in calculus for finding the derivative of composite functions. Let's break it down simply: suppose you have a function within another function, like an onion with multiple layers.
Imagine a sandwich where the meat (inner function) is wrapped by the bread (outer function). To find the rate of change (derivative) of this entire sandwich (composite function), you need to determine the derivative of the outer layer first while keeping the inner layer unchanged.
Then, you find the derivative of the inner layer itself. Finally, you multiply these two derivatives together. This approach allows you to effectively "peel back the layers" of the sandwich (or onion) to understand how the entire function changes.
  • The outer layer in this example is: \(u^{1/3}\)
  • The inner layer is: \(u = 2r - r^2\)
  • The result is: Multiply the derivatives of the outer and inner functions.
Differentiation of Composite Functions
When differentiating composite functions, you apply the chain rule, as discussed. Composite functions are functions made up of more than one simpler function.
By differentiating each part of the function layer by layer, you start by focusing on the most outside layer, gradually working your way inward.
In practice, you apply the rule: identify the roles of each function - the inner and the outer.
Then, differentiate the outer function first while holding the inner function constant. Once you've done that, differentiate the inner function.
For the exercise at hand, this involves recognizing that \( (2r - r^2)^{1/3} \) contains the inner function \(u = 2r - r^2\) and the outer function \(u^{1/3}\). These two roles precisely illustrate how composite functions are differentiated step by step by peeling each layer away using the chain rule.
Power Rule
The power rule is one of the fundamental rules in calculus, used to differentiate functions of the form \(x^n\). According to the power rule, you bring down the exponent as a multiplier and subtract 1 from the original exponent.
So, the formula for the derivative of \(x^n\) is \(nx^{n-1}\).
This rule simplifies the process of finding derivatives and is particularly handy when dealing with polynomial functions.
In the given problem, after rewriting the function to \( (2r - r^2)^{1/3} \), the power rule helps us find the derivative of the outer function. Here, \(n = 1/3\), therefore the derivative of the outer part becomes \((1/3)(u^{-2/3})\).
It's a quick way to tackle powers and makes complex differentiation tasks more manageable, working especially well when combined with other rules like the chain rule.

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Most popular questions from this chapter

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