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Chapter 3: Problem 14

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

Short Answer

Expert verified
Answer: The derivative of the function \(s(t) = 4e^t\sqrt{t}\) with respect to \(t\) is \(\frac{ds}{dt} = 4e^t\sqrt{t}\left(1 + \frac{1}{2t}\right)\).

Step by step solution

01

Identify the functions involved in the product

In this case, we see that the given function can be expressed as a product of two functions: \(f(t)=4e^t\) and \(g(t)=\sqrt{t}\), so \(s(t)=f(t)g(t)\).
02

Use the product rule to find the derivative

For two functions \(f(t)\) and \(g(t)\), the product rule states that the derivative of their product is given by: \[\frac{d}{dt}(f(t)g(t))=f'(t)g(t)+f(t)g'(t)\] We'll find derivatives of each function separately and then apply the product rule.
03

Differentiate \(f(t)=4e^t\)

The derivative of an exponential function is straightforward: \[\frac{d}{dt}(4e^t)=4e^t\] So, \(f'(t)=4e^t\).
04

Differentiate \(g(t)=\sqrt{t}\)

First, rewrite \(\sqrt{t}\) as \(t^{\frac{1}{2}}\). Now, apply the chain rule: \[\frac{d}{dt}(t^{\frac{1}{2}})=\frac{1}{2}t^{\frac{1}{2}-1} = \frac{1}{2}t^{-\frac{1}{2}}\] So, \(g'(t)=\frac{1}{2}t^{-\frac{1}{2}}\).
05

Apply the product rule

Now that we have the derivatives for the individual functions, we can apply the product rule: \[\frac{d}{dt}(s(t))=f'(t)g(t)+f(t)g'(t)=4e^t\sqrt{t}+4e^t\cdot\frac{1}{2}t^{-\frac{1}{2}}\]
06

Simplify the expression and write the final answer

Combine the terms in the expression: \[\frac{d}{dt}(s(t))=4e^t\sqrt{t}\left(1 + \frac{1}{2t}\right)\] The derivative of the function \(s(t)=4e^t\sqrt{t}\) with respect to \(t\) is: \[\frac{ds}{dt}=4e^t\sqrt{t}\left(1 + \frac{1}{2t}\right)\]

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Most popular questions from this chapter

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