Problem 33 - Single Variable Calculus

Question

Find the derivative of the function and \(\underline{\text { eval }}\) uate \(f^{\prime}(x)\) at the given value of \(x\). \(f(x)=(\sqrt{x}+2 x)\left(x^{3 / 2}-x\right) ; \quad x=4\)

Step-by-Step Solution

Answer

\(f'(x) = (\frac{1}{2}x^{1/2})(x^{3/2} - x) + 2(x^{3/2} - x) + \frac{3}{2}x( \sqrt{x} + 2x) - (\sqrt{x} + 2x)\) Evaluating at \(x=4\), we get \(f'(4) = -20\).

Step 1: Identify the functions

First, we will identify the functions that need to be differentiated. In this exercise, we have: - \(f(x) = \sqrt{x} + 2x\) - \(g(x) = x^{3/2} - x\)

Step 2: Compute the derivatives of f(x) and g(x)

We will compute the derivatives of \(f(x)\) and \(g(x)\) independently using the power rule for differentiation. - \(f'(x) = \frac{d}{dx} (\sqrt{x} + 2x) = \frac{1}{2}x^{-1/2} + 2\) - \(g'(x) = \frac{d}{dx} (x^{3/2} - x) = \frac{3}{2}x^{1/2} - 1\)

Step 3: Apply the product rule

Now, we can apply the product rule for differentiation to find the derivative of the given function \(f(x)g(x)\). So, \(f'(x)g(x) + f(x)g'(x) = (\frac{1}{2}x^{-1/2} + 2)(x^{3/2} - x) + (\sqrt{x} + 2x)(\frac{3}{2}x^{1/2} - 1)\)

Step 4: Simplify the expression

Simplify the expression to get a more manageable form of the derivative. \(f'(x)g(x) + f(x)g'(x) = (\frac{1}{2}x^{1/2})(x^{3/2} - x) + 2(x^{3/2} - x) + \frac{3}{2}x( \sqrt{x} + 2x) - (\sqrt{x} + 2x)\)

Step 5: Evaluate the derivative at x = 4

Now, we can evaluate the derivative at x = 4. \(f'(4)g(4) + f(4)g'(4) = (\frac{1}{2} \cdot 4^{1/2})(4^{3/2} - 4) + 2(4^{3/2} - 4) + \frac{3}{2} \cdot 4( 4^{1/2} + 2 \cdot 4) - (4^{1/2} + 2 \cdot 4)\) After evaluating the above expression, we get: \(f'(4) = -20\) So, the derivative of the function at x = 4 is \(-20\).

Key Concepts

Product Rule for Differentiation

When we have to find the derivative of a function that is the product of two or more functions—like in our given exercise, where we have to differentiate the product of \(f(x)\) and \(g(x)\)—we use the product rule for differentiation. This rule states that the derivative of a product is the derivative of the first function times the second function plus the first function times the derivative of the second function.

In formula terms, if we have two functions \( u(x) \) and \( v(x)\), their derivative \( u'(x)v(x) + u(x)v'(x) \) is the application of the product rule. In the exercise, we applied the product rule to \(f(x)\) and \(g(x)\) after determining their individual derivatives using the power rule.

Power Rule for Differentiation

The power rule is one of the most fundamental rules for taking derivatives, especially when the function is a polynomial or has terms with powers of \(x\). It says that if \(f(x) = x^n\), where \(n\) is a constant, then the derivative \(f'(x) = nx^{n-1}\).

In the context of our exercise, we used the power rule to differentiate \(\sqrt{x}\), which is \(x^{1/2}\), and the terms in \(x^{3/2}\) and \(x\). This gave us \(f'(x)\) and \(g'(x)\) conveniently, which we then used in the product rule to find the derivative of the entire expression.

Evaluating Derivatives

After finding the derivative of a function, we often need to evaluate it at a specific point to understand the rate of change at that point. This involves plugging the value of \(x\) into our derivative formula. As with our exercise, evaluating derivatives is crucial in applications such as finding the slope of a tangent line to a curve at a certain point or determining the instantaneous rate of change.

For our exercise, we evaluated the derivative at \(x=4\), which required substituting \(4\) into our derivative expression and simplifying to find that \(f'(4) = -20\). This tells us that at \(x=4\), the rate of change of our function is \( -20 \) units.

Simplifying Expressions

Simplifying expressions is an essential step in calculus to make the result more understandable and easier to work with. In the context of differentiation, after applying rules like the product and power rule, we often end up with complex expressions that can be simplified by combining like terms, factoring, expanding, and canceling where appropriate.

In the provided exercise, after applying the product rule, we were left with a derivative that needed simplification for evaluation. This step is not just about making the expression look neater; it often reveals the function's characteristics and behaviors more clearly and makes further computations less prone to error.