Problem 14 - Calculus Early Transcendentals

Question

Find the derivative of the following functions. $$s(t)=4 e^{t} \sqrt{t}$$

Step-by-Step Solution

Answer

Question: Find the derivative of the function $$s(t) = 4e^t\sqrt{t}$$. Answer: The derivative of the function $$s(t) = 4e^t\sqrt{t}$$ is $$s'(t) = 4e^t\sqrt{t} + \frac{2e^t}{\sqrt{t}}$$.

Step 1: Identify the two functions

In the given function, $$s(t) = 4e^t\sqrt{t}$$, we have two functions multiplied together: $$f(t) = 4e^t$$ $$g(t) = \sqrt{t}$$

Step 2: Find the derivative of f(t)

The function $$f(t) = 4e^t$$ is a simple exponential function. To find its derivative, we apply the following rule: $$(e^t)' = e^t$$ So, the derivative of $$f(t)$$ is: $$f'(t) = 4 \cdot e^t$$

Step 3: Find the derivative of g(t)

The function $$g(t) = \sqrt{t}$$ can be rewritten as: $$g(t) = t^{\frac{1}{2}}$$ To find its derivative, we apply the power rule, which states that if $$g(t) = t^n$$, then $$g'(t) = n \cdot t^{n-1}$$ So, the derivative of $$g(t)$$ is: $$g'(t) = \frac{1}{2} \cdot t^{-\frac{1}{2}}$$

Step 4: Apply the product rule

Now that we have the derivatives of both functions, we can apply the product rule to find the derivative of $$s(t)$$: $$s'(t) = f'(t)g(t) + f(t)g'(t)$$ Substitute the derivatives of $$f(t)$$ and $$g(t)$$ we found earlier to get: $$s'(t) = (4 \cdot e^t)(\sqrt{t}) + (4e^t)\left(\frac{1}{2} \cdot t^{-\frac{1}{2}}\right)$$

Step 5: Simplify the expression

Now we can simplify the expression for $$s'(t)$$: $$s'(t) = 4e^t\sqrt{t} + 2e^tt^{-\frac{1}{2}}$$ $$s'(t) = 4e^t\sqrt{t} + \frac{2e^t}{\sqrt{t}}$$ The derivative of the function $$s(t) = 4e^t\sqrt{t}$$ is: $$s'(t) = 4e^t\sqrt{t} + \frac{2e^t}{\sqrt{t}}$$

Key Concepts

Product Rule

When you're dealing with the derivative of a product of two functions, the product rule is your best friend. This rule helps you find the derivative of two functions that are multiplied together. If you have two functions, say \( f(t) \) and \( g(t) \), the product rule states that the derivative of their product is given by:

• \( (fg)' = f'g + fg' \)

Here's how it works:

• Differentiate the first function, \( f(t) \), and keep the second function, \( g(t) \), unchanged.
• Then, add the product of the unchanged first function with the derivative of the second function, \( g(t) \).

In our exercise with \( s(t) = 4e^t\sqrt{t} \), the functions \( f(t) = 4e^t \) and \( g(t) = \sqrt{t} \) are multiplied. We used the product rule to find \( s'(t) \) by adding \( f'(t)g(t) \) with \( f(t)g'(t) \). This gave us the derivative of \( s(t) \) by solving each part step by step, ensuring clarity in solving.

Exponential Functions

Exponential functions are very particular in calculus due to their straightforward derivative properties. If you have a function of the form \( e^t \), the derivative is exceptionally simple:

• \( (e^t)' = e^t \)

This means that no matter the input value, the rate of change of an exponential function \( e^t \) remains the same as its value. It's like magic because the derivative is the same as the original function's value!
In the given exercise, \( f(t) = 4e^t \) is part of our original function \( s(t) \). Here, the constant 4 is carried along when differentiating, resulting in \( f'(t) = 4e^t \). This smooth process is why exponential functions are often seen as the "easy part" of differentiation.

Power Rule

The power rule is one of the fundamental rules in differentiation and is especially useful for dealing with polynomial functions and functions with powers. It states that if you have a function \( t^n \), its derivative is:

• \( (t^n)' = n \cdot t^{n-1} \)

This means you multiply by the power and reduce the power by one.
For the function \( g(t) = \sqrt{t} \), we treat it as a power function \( t^{1/2} \). Applying the power rule:

• We get \( g'(t) = \frac{1}{2} \cdot t^{-1/2} \)

With this power rule application, we simplified the derivative process, allowing us to incorporate it easily into the overall product rule application, leading to our final derivative result of the function \( s(t) = 4e^t\sqrt{t} \). The power rule simplifies the seemingly complex derivatives into manageable steps.