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Chapter 3: Problem 23

Derivatives Find and simplify the derivative of the following functions. $$f(t)=t^{5 / 3} e^{t}$$

Short Answer

Expert verified
Question: Find the derivative of the function $$f(t) = t^{\frac{5}{3}} e^t$$ with respect to t and simplify the result. Answer: The derivative of the given function is $$h'(t) = t^{\frac{2}{3}}e^t (\frac{5}{3} + t)$$.

Step by step solution

01

Find the derivative of $$f(t)$$ and $$g(t)$$ separately.

We will begin by finding the derivatives of $$f(t)$$ and $$g(t)$$ individually. For the function, $$f(t) = t^{\frac{5}{3}}$$, its derivative, $$f'(t)$$, can be found using the power rule, which states that the derivative of $$t^n$$, where n is a constant, is given by $$nt^{n-1}$$. Thus, the derivative of $$f(t)$$ is: $$f'(t) = (\frac{5}{3})t^{\frac{5}{3} - 1} = \frac{5}{3}t^{\frac{2}{3}}$$ For the function, $$g(t) = e^t$$, its derivative, $$g'(t)$$, is also $$e^t$$, since the derivative of $$e^t$$ with respect to t is itself. Now we have the derivatives of both functions: $$f'(t) = \frac{5}{3}t^{\frac{2}{3}}$$ $$g'(t) = e^t$$
02

Apply the product rule to find the derivative of $$h(t)$$.

We will now apply the product rule to find the derivative of the given function, $$h(t) = f(t)g(t)$$. According to the product rule, the derivative of h(t) is: $$h'(t) = f'(t)g(t) + f(t)g'(t)$$ Now substitute the derivatives and original functions into the equation: $$h'(t) = (\frac{5}{3}t^{\frac{2}{3}})(e^t) + (t^{\frac{5}{3}})(e^t)$$
03

Simplify the derivative.

Now that we have substituted our functions into the product rule equation, we can simplify it by combining like terms: $$h'(t) = (\frac{5}{3}t^{\frac{2}{3}}e^t + t^{\frac{5}{3}}e^t)$$ Factor out the common term of $$t^{\frac{2}{3}}e^t$$: $$h'(t) = t^{\frac{2}{3}}e^t (\frac{5}{3} + t)$$ We now have the simplified derivative of $$f(t) = t^{\frac{5}{3}} e^t$$: $$h'(t) = t^{\frac{2}{3}}e^t (\frac{5}{3} + t)$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
Understanding the product rule is essential when dealing with derivatives of functions that are multiples of each other. In calculus, the product rule is used to find the derivative of the product of two functions. The rule states that if you have two differentiable functions, let's say f(t) and g(t), the derivative of their product h(t) = f(t)g(t) is given by the formula:

\[ h'(t) = f'(t)g(t) + f(t)g'(t) \]
  • Firstly, find the derivative of each function separately.
  • Next, multiply the first function's derivative by the second function.
  • Then, multiply the first function by the derivative of the second function.
  • Finally, add the two products together to get the derivative of the product of the two functions.
This allows the differentiation of more complex expressions by breaking them down into simpler parts.
Power Rule
The power rule is a quick method for finding the derivative of a function that is a power of t. It states that if you have a function f(t) = t^n, where n is a real number, the derivative of that function with respect to t is given by multiplying the power by the coefficient and then subtracting one from the power:

\[ f'(t) = nt^{n-1} \]
  • Identify the power n of the function.
  • Multiply the function by n.
  • Reduce the power of t by 1 to find the new exponent.
The simplicity of the power rule makes it one of the most commonly used techniques in differentiating polynomial functions.
Derivative Simplification
Once the derivatives have been found using the product rule, power rule, or other differentiation methods, it's often necessary to simplify the expression to its most basic form. This process makes the result easier to interpret and may be essential for further calculations. To simplify a derivative:

  • Combine like terms by adding or subtracting coefficients.
  • Factor out any common factors from terms.
  • Reduce fractions if possible to simplify the expression.
Derivative simplification can involve algebraic manipulation such as expanding products, collecting like terms, and factoring. In our example, factoring out the common term t^{2/3}e^t simplifies the expression considerably, making it more concise and manageable.
Exponential Functions
Exponential functions are functions of the form f(t) = a^t, where a is a constant. The natural exponential function, where a = e (Euler's number, approximately 2.718), is particularly important in calculus because of its unique properties. One of these properties is that the function's derivative is the same as the function itself. Thus:

\[ \frac{d}{dt}(e^t) = e^t \]
  • The derivative of e^t with respect to t is simply e^t.
  • Exponential functions model growth and decay processes in nature and finance.
  • Their constant derivative makes them particularly useful in solving differential equations.
Remembering this property simplifies finding derivatives involving exponential terms, as seen in our example.

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