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Find the first and second derivatives of the function. \(G(r)=\sqrt{r}+\sqrt[3]{r}\)

Short Answer

Expert verified
The first derivative is \( G'(r) = \frac{1}{2\sqrt{r}} + \frac{1}{3\sqrt[3]{r^2}} \) and the second derivative is \( G''(r) = -\frac{1}{4r^{3/2}} - \frac{2}{9r^{5/3}} \).

Step by step solution

01

Rewrite the Function

To make differentiation easier, rewrite each term of the function using exponents. Remember that \(\sqrt{r}\) is the same as \(r^{1/2}\) and \(\sqrt[3]{r}\) is the same as \(r^{1/3}\). So, \(G(r) = \sqrt{r} + \sqrt[3]{r}\) can be written as \(G(r) = r^{1/2} + r^{1/3}\).
02

Differentiate the Function (First Derivative)

Now, find the derivative of \(G(r) = r^{1/2} + r^{1/3}\). Use the power rule, which states that the derivative of \(r^n\) is \(n \cdot r^{n-1}\). The first derivative is: \[ G'(r) = \frac{1}{2} r^{-1/2} + \frac{1}{3} r^{-2/3} \]
03

Simplify the First Derivative

Simplify \(G'(r)\) by expressing it using radicals again. This gives: \[ G'(r) = \frac{1}{2\sqrt{r}} + \frac{1}{3\sqrt[3]{r^2}} \]
04

Differentiate Again (Second Derivative)

Now, find the second derivative of \(G(r)\) by differentiating \(G'(r) = \frac{1}{2} r^{-1/2} + \frac{1}{3} r^{-2/3}\) using the power rule again. Second derivative:\[ G''(r) = -\frac{1}{4} r^{-3/2} - \frac{2}{9} r^{-5/3} \]
05

Simplify the Second Derivative

Simplify \(G''(r)\) by expressing it using radicals: \[ G''(r) = -\frac{1}{4r^{3/2}} - \frac{2}{9r^{5/3}} \]

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Key Concepts

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

Power Rule
One of the most fundamental rules in calculus for finding derivatives is the Power Rule. It's your go-to formula when you encounter expressions with exponents. Suppose you have a function of the form
  • \( f(x) = x^n \),
the Power Rule states that the derivative \( f'(x) \) is
  • \( f'(x) = n \cdot x^{n-1} \).
This simple rule allows you to quickly find the derivative of any power of \( r \). In the exercise at hand, the function \( G(r) = r^{1/2} + r^{1/3} \) fits perfectly into this framework. You apply the Power Rule to each term separately:
  • For \( r^{1/2} \), the derivative becomes \( \frac{1}{2} r^{-1/2} \).
  • For \( r^{1/3} \), the derivative becomes \( \frac{1}{3} r^{-2/3} \).
Using the Power Rule in both steps, you can differentiate with ease.
Simplifying Derivatives
After applying the Power Rule, the resulting derivatives are often expressed with negative exponents. This can make them look complex, especially if you want to evaluate them further. To make these derivatives more manageable, it's helpful to transform them back into a form involving radicals.
  • For instance, \( r^{-1/2} \) becomes \( \frac{1}{\sqrt{r}} \) because \( \sqrt{r} \) is equivalent to \( r^{1/2} \).
  • Similarly, \( r^{-2/3} \) transforms into \( \frac{1}{\sqrt[3]{r^2}} \). This is because \( \sqrt[3]{r^2} \) corresponds to \( r^{2/3} \).
Simplifying derivatives not only makes them look cleaner. It also makes evaluating these expressions at specific values of \( r \) easier. By rewriting derivatives with radicals, you stay in familiar territory. This is especially true if you've previously learned about dealing with square roots or cube roots in algebra.
Radical Expressions
Radical expressions involve roots, such as square roots and cube roots. They're common in many areas of mathematics, including calculus. When you first look at a function like \( \sqrt{r} + \sqrt[3]{r} \), you might think differentiation is complex. However, by expressing radicals as exponents, it becomes much more manageable.
  • A square root, \( \sqrt{r} \), translates to \( r^{1/2} \) in exponential form.
  • Similarly, a cube root, \( \sqrt[3]{r} \), is written as \( r^{1/3} \).
These transformations unlock the door to using calculus's full power, like the Power Rule for derivatives. Understanding how to shift between radicals and exponents is a valuable skill. It helps in simplifying expressions and solving equations. Additionally, it’s crucial for more advanced mathematics in calculus and beyond. Because it's easier to differentiate using exponents, these transformations allow for a streamlined process in finding and simplifying derivatives.

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