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Find the derivative of the function. \(f(u)=\frac{1}{\sqrt{u}}-\frac{3}{\sqrt[3]{u}}\)

Short Answer

Expert verified
The derivative of the function \(f(u) = \frac{1}{\sqrt{u}} - \frac{3}{\sqrt[3]{u}}\) is: \(f'(u) = -\frac{1}{2\sqrt{u^3}} - \frac{1}{u^{\frac{4}{3}}}\).

Step by step solution

01

1. Rewrite the given function using exponents

First, let's rewrite the two terms of f(u) using exponents instead of radicals. This makes it simpler for us to apply the power rule and find the derivatives. 1. \(\frac{1}{\sqrt{u}}\) can be written as \(u^{-\frac{1}{2}}\) 2. \(\frac{3}{\sqrt[3]{u}}\) can be written as \(3u^{-\frac{1}{3}}\) So now our function becomes: \(f(u) = u^{-\frac{1}{2}} - 3u^{-\frac{1}{3}}\)
02

2. Apply the power rule to determine the derivative of each term

Apply the Power Rule to the given function with exponents: 1. The derivative of \(u^{-\frac{1}{2}}\) is \(f'(u) = -\frac{1}{2}u^{-\frac{3}{2}}\) 2. The derivative of \(3u^{-\frac{1}{3}}\) is \(f'(u) = -u^{-\frac{4}{3}}\) Since the first term doesn't have a constant, we don't need to adjust the coefficient. Now, let's combine the derivatives: \(f'(u) = -\frac{1}{2}u^{-\frac{3}{2}} - u^{-\frac{4}{3}}\)
03

3. Rewrite the derivatives as radicals and simplify

Now that we have the derivatives with exponents, let's rewrite them using radicals: 1. \(-\frac{1}{2} u^{\left(-\frac{3}{2}\right)}\) = \(-\frac{1}{2} \frac{1}{\sqrt{u^3}}\) = \(-\frac{1}{2\sqrt{u^3}}\) 2. \(-u^{\left(-\frac{4}{3}\right)}\) = \(-\frac{1}{\sqrt[3]{u^4}}\) = \(-\frac{1}{u^{\frac{4}{3}}}\) Now, we have the derivatives written using radicals: \(f'(u) = -\frac{1}{2\sqrt{u^3}} - \frac{1}{u^{\frac{4}{3}}}\)
04

Final Answer

We found the derivatives of both terms, combined them and rewrote it using radicals. So the derivative of the given function f(u) is: \(f'(u) = -\frac{1}{2\sqrt{u^3}} - \frac{1}{u^{\frac{4}{3}}}\)

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Key Concepts

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

Power Rule
When it comes to finding derivatives in calculus, the power rule is a fundamental tool that simplifies the process tremendously. It states that if you have a function in the form of f(x) = x^n, where n is any real number, the derivative of f(x) with respect to x is given by f'(x) = nx^{n-1}. This rule applies no matter the value of n, whether it is positive, negative, or even a fraction.

In the provided exercise, using the power rule made it manageable to find the derivative of terms involving exponents and radicals. It's crucial to understand that rewriting the function with exponents was an important step before applying the power rule. This rule drastically reduced the complexity of the problem, turning the daunting task of differentiating radicals into a much more approachable one using simple exponent manipulation.
Exponents and Radicals
Exponents and radicals are two sides of the same coin; they are mathematical expressions related to the operation of raising a number to a power. An exponent indicates how many times a number, known as the base, is multiplied by itself. A radical, or a root, reversely expresses the process of finding a number that, when raised to the specified power (the index of the radical), results in the given value.

Understanding the relationship between exponents and radicals is key in simplifying functions for differentiation. In the original exercise, radicals were converted into exponent form, where the nth root of x translates to x^(1/n). This translation to exponent form is crucial for applying the power rule in calculus and for the simplification of a function into a form where the derivative can be more easily calculated.
Derivative of a Function
The derivative of a function at a point is the rate at which the function value changes with respect to change in its input value. It is a core concept in calculus, providing a mathematical way to discuss and calculate instantaneous rates of change and slopes of tangents to curves. In practice, finding the derivative involves taking functions that may have one or several terms and applying rules like the power rule, product rule, quotient rule, and chain rule to get the derivative of the function.

For the exercise on hand, understanding how to find the derivative of a function allowed us to transform each term of the given function separately and then combine them to find the complete derivative. The step-by-step solution walked through differentiating two terms, highlighting that solving for the derivative requires attention to each component of the equation.
Calculus
Calculus is a branch of mathematics that deals with rates of change (differential calculus) and accumulation of quantities (integral calculus). It's a powerful tool that has applications ranging from physics to engineering, economics, and beyond. Calculus can model and solve problems involving complex dynamic systems, providing insights into motion, growth, decay, and an array of natural phenomena.

In this context, calculus allows us to dissect functions, like the one in the exercise, into their derivatives or integrals—giving us insight into the instantaneous behavior of functions and enabling us to solve for areas under curves, respectively. Approaching problems in calculus often follows a set pattern of identifying the type of function, picking the right rules, and manipulating the equation to find the solution, all of which were employed in finding the derivative for this particular example.

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Most popular questions from this chapter

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