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Chapter 4: Problem 17

Differentiate the following functions. \(f(t)=\frac{e^{3 t}+e^{-3 t}}{e^{t}}\)

Short Answer

Expert verified
The derivative of \(f(t)\) is \(2e^{2t} - 4e^{-4t}\).

Step by step solution

01

Simplify the Function

Rewrite the function to make differentiation easier. Given the function is \[f(t)=\frac{e^{3 t}+e^{-3 t}}{e^{t}}\]divide each term in the numerator by the denominator: \[f(t)=\frac{e^{3 t}}{e^{t}} + \frac{e^{-3 t}}{e^{t}} = e^{2 t} + e^{-4 t}\]
02

Differentiate Each Term

Differentiate each term with respect to t. For the first term, use the rule \(\frac{d}{dt}[e^{u}] = e^{u} \cdot \frac{du}{dt}\). Hence,\[\frac{d}{dt}[e^{2t}] = 2e^{2t}\]For the second term, use the same rule:\[\frac{d}{dt}[e^{-4t}] = e^{-4t} \cdot (-4) = -4e^{-4t}\]
03

Combine the Results

Combine the results of the differentiation to get the final derivative:\[f'(t) = 2e^{2t} - 4e^{-4t}\]

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Key Concepts

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

derivative rules
When differentiating functions, it鈥檚 essential to understand and properly apply the basic rules. Here are some essential derivative rules you need to know:
  • Power Rule: If you have a function of the form \( f(x) = x^n \), the derivative is \( f'(x) = nx^{n-1} \).
  • Product Rule: To differentiate a product of two functions, use \( (uv)' = u'v + uv' \).
  • Quotient Rule: To differentiate a quotient, apply \( \bigg( \frac{u}{v} \bigg)' = \frac{u'v - uv'}{v^2} \).
  • Chain Rule: For a composition of functions, \( f(g(x)) \), use \( (f \big( g(x) \big))' = f' \big( g(x) \big) \times g'(x) \).
In our problem, we use the rule for exponential functions: \( \frac{d}{dt}[e^{u}] = e^{u} \times \frac{du}{dt} \). Applying it helps in differentiating exponential terms easily.
simplifying functions
Simplifying the function is a crucial step before differentiation. Simplification makes processes easier and less error-prone.

Look at the initial function, \( f(t)=\frac{e^{3 t}+e^{-3 t}}{e^{t}} \). Instead of directly differentiating, simplify it by breaking it down:
  • Divide each term in the numerator by \( e^{t} \): \( \frac{e^{3 t}}{e^{t}} + \frac{e^{-3 t}}{e^{t}} \).
  • This transforms to \( e^{2 t} + e^{-4 t} \), a more manageable form.
By simplifying first, differentiation becomes straightforward. Always look to simplify functions where possible to streamline the process.
exponential functions
Exponential functions are pervasive in calculus and appear in many forms. Let鈥檚 breakdown how to handle them:

Consider a term like \( e^{u} \). Its derivative involves applying the chain rule in most cases:\( \frac{d}{dt}[e^{u}] = e^{u} \times \frac{du}{dt} \). Here鈥檚 how it works practically:
  • For \( e^{2t} \), \( u = 2t \). Thus, \( \frac{du}{dt} = 2 \), so \( \frac{d}{dt}[e^{2t}] = 2e^{2t} \).
  • For \( e^{-4t} \), \( u = -4t \). Thus, \( \frac{du}{dt} = -4 \), so \( \frac{d}{dt}[e^{-4t}] = -4e^{-4t} \).
By combining these individual derivatives, we obtain the final result for the derivative:\( f'(t) = 2e^{2t} - 4e^{-4t} \).

Remember, understanding and applying these rules accurately is key to mastering differentiation of exponential functions.

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