Question

Differentiate the following functions: \(u=\frac{1-x^{-\frac{1}{2}}}{x^{\frac{1}{3}}}\)

Short answer

1. Find the derivatives of the numerator and the denominator functions: \(f'(x)=-\frac{1}{2}(x)^{-\frac{3}{2}}\) and \(g'(x)=\frac{1}{3}(x)^{-\frac{2}{3}}\). 2. Apply the quotient rule: \(u'(x)=\frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\) 3. Substitute the values of \(f'(x)\), \(g'(x)\), \(f(x)\), and \(g(x)\) in the quotient rule formula. 4. Simplify the expression to obtain the derivative: \(u'(x) = \frac{-\frac{1}{2}x^{-\frac{7}{6}} - \frac{1}{3}x^{-\frac{5}{6}}+\frac{1}{3}(x)^{-1}}{x^{\frac{2}{3}}}\).

Step-by-step solution

Find \(f'(x)\)

To find \(f'(x)\), we will need to differentiate \(f(x) = 1-x^{-\frac{1}{2}}\). Using the power rule ( \(\frac{d}{dx}(x^n)=nx^{n-1}\) ), we have: \(f'(x)=-\frac{1}{2}(x)^{-\frac{3}{2}}\)

Find \(g'(x)\)

To find \(g'(x)\), we will need to differentiate \(g(x) = x^{\frac{1}{3}}\). Using the power rule ( \(\frac{d}{dx}(x^n)=nx^{n-1}\) ), we have: \(g'(x)=\frac{1}{3}(x)^{-\frac{2}{3}}\)

Apply the quotient rule

Now, we will apply the quotient rule to find the derivative of the function \(u(x)\). Using the results from steps 1 and 2, we have: \(u'(x)=\frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\) \(u'(x)=\frac{(-\frac{1}{2}(x)^{-\frac{3}{2}})(x^{\frac{1}{3}}) - (1-x^{-\frac{1}{2}})\left(\frac{1}{3}(x)^{-\frac{2}{3}}\right)}{(x^{\frac{1}{3}})^2}\) \(u'(x)=\frac{-\frac{1}{2}x^{-\frac{3}{2}+\frac{1}{3}} - \frac{1}{3}x^{-\frac{1}{2}-\frac{2}{3}}+\frac{1}{3}(x)^{-1}}{x^{\frac{2}{3}}}\)

Simplify the expression

To simplify the expression, we can simply add and subtract the exponents inside the derivative: \(u'(x)=\frac{-\frac{1}{2}x^{-\frac{3}{2}+\frac{1}{3}} - \frac{1}{3}x^{-\frac{1}{2}-\frac{2}{3}}+\frac{1}{3}(x)^{-1}}{x^{\frac{2}{3}}} = \frac{-\frac{1}{2}x^{-\frac{7}{6}} - \frac{1}{3}x^{-\frac{5}{6}}+\frac{1}{3}(x)^{-1}}{x^{\frac{2}{3}}}\) So, the derivative of the function \(u(x)\) is: \(u'(x) = \frac{-\frac{1}{2}x^{-\frac{7}{6}} - \frac{1}{3}x^{-\frac{5}{6}}+\frac{1}{3}(x)^{-1}}{x^{\frac{2}{3}}}\).

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus where we determine the rate at which a function changes at any given point. It is the process of calculating a derivative, which is the slope of the tangent line to the curve of a function at a certain point. In essence, the derivative measures how sensitive the output value of a function is to changes in its input value.

In practical terms, finding a derivative can help us understand the rate of change in real-world scenarios, such as the acceleration of a car when its velocity is changing. For example, if distance over time is represented as a function, the derivative gives us the velocity. When we take the derivative of a velocity function, we get the acceleration. This cascading effect showcases how derivatives help us quantify change in various contexts.

Quotient Rule

The quotient rule is a technique used in differentiation when dealing with the division of two functions. Specifically, if you have a function which can be expressed as the quotient of two other functions, say, \(u = \frac{f(x)}{g(x)}\), then the derivative of \(u\), denoted \(u'(x)\), is given by: \(u'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\).

This formula is derived from the limit definition of a derivative and is a convenient rule to remember when faced with complex fractional functions. The quotient rule is instrumental in ensuring that we can accurately calculate derivatives without resorting to the first principles for each new problem we encounter.

Power Rule

The power rule simplifies the process of taking derivatives of monomial functions, which are functions where the variable \(x\) is raised to a power \(n\). According to the power rule, if \( f(x) = x^n \) for any real number \( n \) (except \( n = 0 \)), then the derivative of \( f(x)\) with respect to \( x \) is \( f'(x) = nx^{n-1}\).

For instance, if we have a function \( g(x) = x^5 \) and we use the power rule, the derivative is \( g'(x) = 5x^{5-1} = 5x^4\). This rule greatly expedites finding derivatives of functions that would otherwise require the application of the limit definition, which can be much more time-consuming.

Simplifying Expressions

Simplifying expressions in calculus is often the final step after applying various rules of differentiation like the quotient and power rules. This step is crucial for ensuring that solutions are presented in the most accessible form. Simplification can involve combining like terms, reducing fractions, and eliminating complex expressions.

For example, in solving our initial problem, we simplified \( u'(x) \) by combining terms with common exponents and cancelling out factors where possible. This process not only makes the derivative easier to interpret but also aids in further applications, such as solving equations that the derivative may be a part of. Simplifying expressions is an important skill, fostering clear communication and efficiency in solving mathematical problems.