Problem 32 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the derivative of the function \(f\) by using the rules of differentiation. \(f(t)=2 t^{2}+\sqrt{t^{3}}\)

Step-by-Step Solution

Answer

The derivative of the given function is \(f'(t) = 4t + \frac{3}{2}t^{\frac{3}{2}}\).

Step 1: 1. Apply the Power Rule

We must apply the Power Rule to differentiate the function with respect to t. The Power Rule states that if \(f(t) = t^n\), then \(f'(t) = nt^{(n-1)}\). In this case, \(f(t) = 2t^2\). So to find the derivative, we apply the Power Rule: \( f'(t) = 2(2t^{(2-1)}) \)

Step 2: 2. Differentiate the Square Root Function

Next, we need to differentiate the square root function. To do this, let's rewrite the square root function as a power function: \( \sqrt{t^3} = (t^3)^{\frac{1}{2}} \) Now, we can apply the Power Rule and the Chain Rule. The Chain Rule states that if \(f(t) = g(h(t))\), then \(f'(t) = g'(h(t))\cdot h'(t)\). In this case, \(g(t) = t^{1/2}\), and \(h(t) = t^3\). So, we can find the derivatives of \(g(t)\) and \(h(t)\): \(g'(t) = \frac{1}{2}(t^{(\frac{1}{2}-1)}) = \frac{1}{2}(t^{-\frac{1}{2}}) \) \(h'(t) = 3(t^{(3-1)}) = 3t^2 \) Now, we apply the Chain Rule for \(f'(t) = g'(h(t))\cdot h'(t)\): \( f'(t) = (\frac{1}{2}(t^{3})^{-\frac{1}{2}}) \cdot (3t^2) \)

Step 3: 3. Combine the Derivatives

Now we need to combine the derivatives we found from Steps 1 and 2, by adding them together. \( f'(t) = 2(2t^{(2-1)}) + (\frac{1}{2}(t^3)^{-\frac{1}{2}}) (3t^2) \) Simplify the expression: \( f'(t) = 4t + \frac{3}{2}t^{\frac{3}{2}} \)

Step 4: 4. Final Answer

The derivative of the given function is: \( f'(t) = 4t + \frac{3}{2}t^{\frac{3}{2}} \)

Key Concepts

Power Rule

The Power Rule is one of the basic rules of calculus differentiation and makes finding derivatives straightforward. It applies to functions that are in the form of a power, where the base is the variable and the exponent is a real number. This rule states that if you have a function of the form \(f(t) = t^n\), the derivative, denoted as \(f'(t)\), is calculated as \(nt^{(n-1)}\). This means you simply multiply by the exponent and reduce the exponent by one.

For example, in the function \(f(t) = 2t^2\), using the Power Rule involves multiplying the coefficient by the exponent \(2 \times 2\), yielding \(4\), and then subtracting one from the exponent \((2-1)\), resulting in the derivative \(f'(t) = 4t\).

Remember that the derivative represents the rate of change or the slope of the function at any point \(t\). So, whenever you encounter a power of \(t\), the Power Rule is your go-to technique for differentiation.

Chain Rule

The Chain Rule is essential when differentiating composite functions, or functions within functions. These are functions where one function is applied to the result of another function. The rule provides a way to differentiate these by breaking them down into their constituent parts.

According to the Chain Rule, if you have a composite function \(f(t) = g(h(t))\), the derivative \(f'(t)\) is given by \(g'(h(t)) \cdot h'(t)\). In essence, this means you first find the derivative of the outer function and then multiply it by the derivative of the inner function.

Taking our example of the square root term \(\sqrt{t^3}\), we rewrite this as \((t^3)^{\frac{1}{2}}\), so we have an outer function \(g(t) = t^{\frac{1}{2}}\) and an inner function \(h(t) = t^3\). By finding \(g'(t)\) and \(h'(t)\) separately (\(\frac{1}{2}t^{-\frac{1}{2}}\) and \(3t^2\), respectively), and then applying the Chain Rule, we can differentiate this composite function accurately.

Differentiation Techniques

Differentiation in calculus involves a variety of techniques, each tailored to different types of expressions. Mastering these techniques will enable you to differentiate almost any function with confidence. The basics start with the Power Rule, but also include other fundamental rules like the Product Rule and the Quotient Rule—not covered here, but essential for when multiplicative and divisive operations are involved.

Aside from these powerful rules, algebraic manipulations often simplify differentiation. For instance, rewriting a square root using exponents, as shown in our original exercise \(\sqrt{t^3}\), transforms it into a form where the Power Rule can be effectively applied. Moreover, combining techniques is common, as in our exercise where the Chain Rule follows the Power Rule to differentiate more complex functions.

Finally, merging derived components is vital for composite functions. In our example, after separately differentiating each part, we add the results to get the complete derivative. By honing these techniques, students can enhance their analytical skills, enabling efficient and accurate handling of various functions.