Problem 43
Question
Finding a Derivative In Exercises \(33-54,\) find the derivative. $$ g(t)=\left(e^{-t}+e^{t}\right)^{3} $$
Step-by-Step Solution
Verified Answer
The derivative of the function \(g(t) = (e^{-t} + e^{t})^3 \) is \(g'(t) = 3(e^{-t} + e^{t})^2 (-e^{-t} + e^{t})\).
1Step 1: Identify the inner and outer functions
The function \(g(t)\) is a composite function with an outer function \((u)^3\) and an inner function \(u = e^{-t} + e^{t}\). So we need to find the derivatives of the outer function (denote as \(u'\)) and of the inner function (denote as \(v'\)).
2Step 3: Differentiate the inner function
The inner function \(u(t)\) is \(e^{-t} + e^{t}\). The derivative of \(e^x\) is \(e^x\) itself and, following the chain rule, the derivative of \(e^{-t}\) would be \(-e^{-t}\). Thus the total derivative becomes: \(v' = -e^{-t} + e^{t}\).
3Step 4: Apply the chain rule to get the derivative of the original function
Applying the chain rule (\(f'(u(v)) = u'(v).v'\)) we replace \(u'\) with \(3u^2\) and \(v'\) with \(-e^{-t} + e^{t}\). We then replace \(u\) with the inner function \(e^{-t} + e^{t}\). So, \(g'(t)\) becomes \(g'(t) = 3(e^{-t} + e^{t})^2 (-e^{-t} + e^{t})\).
Key Concepts
Chain RuleComposite FunctionsExponential Functions
Chain Rule
Understanding the chain rule is essential when dealing with composite functions in calculus. It is a formula for computing the derivative of the composition of two or more functions. In simpler terms, if you have a function nested within another function, the chain rule helps you find the derivative of the entire combination.
For instance, if you have a function expressed as \( f(g(x)) \), then the derivative of \( f \) with respect to \( x \), using the chain rule, would be \( f'(g(x)) \times g'(x) \). The crucial first step is to determine the inner function (which we apply first) and the outer function (which comes after).
In our exercise, the composite function is \( (e^{-t} + e^{t})^3 \). Here, the outer function is \( u^3 \), and the inner function is \( u = e^{-t} + e^{t} \). To differentiate this composite function, we differentiate the outer function, then multiply it by the derivative of the inner function. This systematic approach simplifies what might initially seem like a daunting task.
For instance, if you have a function expressed as \( f(g(x)) \), then the derivative of \( f \) with respect to \( x \), using the chain rule, would be \( f'(g(x)) \times g'(x) \). The crucial first step is to determine the inner function (which we apply first) and the outer function (which comes after).
In our exercise, the composite function is \( (e^{-t} + e^{t})^3 \). Here, the outer function is \( u^3 \), and the inner function is \( u = e^{-t} + e^{t} \). To differentiate this composite function, we differentiate the outer function, then multiply it by the derivative of the inner function. This systematic approach simplifies what might initially seem like a daunting task.
Composite Functions
A composite function can be thought of as a combination of two functions, where the output of one function becomes the input of another. Imagine a factory assembly line, where the result from one stage is immediately used in the next. Mathematically, if you have two functions, \( f(x) \) and \( g(x) \), their composition is written as \( f(g(x)) \), and this is read as 'f of g of x'.
The function \( g(t) = (e^{-t} + e^{t})^3 \) from the exercise is a composite function. We see the exponentiation by 3 as the last operation that is performed after the operations inside the parentheses. These nested operations are typical for composite functions, which often require the chain rule for differentiation.
Recognizing the composition of functions is a critical skill in calculus, as it allows you to apply the appropriate rules of differentiation to complex expressions.
The function \( g(t) = (e^{-t} + e^{t})^3 \) from the exercise is a composite function. We see the exponentiation by 3 as the last operation that is performed after the operations inside the parentheses. These nested operations are typical for composite functions, which often require the chain rule for differentiation.
Recognizing the composition of functions is a critical skill in calculus, as it allows you to apply the appropriate rules of differentiation to complex expressions.
Exponential Functions
Exponential functions are powerful tools in mathematics, featuring an exponent that contains a variable. They are denoted as \( a^x \), where \( a \) is a constant, and \( x \) is the variable. For calculus, the most important exponential function is the natural exponential function \( e^x \), where \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
One remarkable property of the natural exponential function is that the derivative of \( e^x \) is \( e^x \) itself. This unique characteristic makes calculations involving \( e^x \) more straightforward. In our exercise, the function includes both \( e^{-t} \) and \( e^{t} \), which, when differentiated, result in \( -e^{-t} \) and \( e^{t} \) respectively. This aligns with the property that the derivative of \( e^x \) with respect to \( x \) is \( e^x \), regardless of whether \( x \) is positive or negative, making the differentiation of these terms relatively easy compared to other functions.
One remarkable property of the natural exponential function is that the derivative of \( e^x \) is \( e^x \) itself. This unique characteristic makes calculations involving \( e^x \) more straightforward. In our exercise, the function includes both \( e^{-t} \) and \( e^{t} \), which, when differentiated, result in \( -e^{-t} \) and \( e^{t} \) respectively. This aligns with the property that the derivative of \( e^x \) with respect to \( x \) is \( e^x \), regardless of whether \( x \) is positive or negative, making the differentiation of these terms relatively easy compared to other functions.
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Problem 43
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