Problem 50
Question
Finding a Derivative In Exercises \(33-54,\) find the derivative. $$ y=\frac{e^{2 x}}{e^{2 x}+1} $$
Step-by-Step Solution
Verified Answer
The derivative of the function \(y=\frac{e^{2 x}}{e^{2 x}+1}\) is \(y'=\frac{2e^{2x}}{(e^{2x} + 1)^2}\)
1Step 1: Identifying the upper and lower functions
First, we recognize the function as a quotient with \(e^{2x}\) in the numerator and \(e^{2x}+1\) in the denominator. Thus, we have \(f(x) = e^{2x}\) and \(g(x) = e^{2x} + 1\).
2Step 2: Calculating the derivatives of the upper and lower functions
The derivative of \(f(x) = e^{2x}\) is \(f'(x) = 2e^{2x}\). The derivative of \(g(x) = e^{2x} + 1\) is \(g'(x) = 2e^{2x}\) since the derivative of \(1\) or any constant is \(0\).
3Step 3: Applying the Quotient Rule
Given that \(f'(x) = 2e^{2x}\), \(f(x) = e^{2x}\), \(g'(x) = 2e^{2x}\) and \(g(x) = e^{2x} + 1\), we can now apply the quotient rule: \(\frac{d}{dx} \frac{f(x)}{g(x)} = \frac{f'(x) g(x) - g'(x) f(x)}{(g(x))^2}= \frac{2e^{2x}(e^{2x} + 1) - 2e^{2x} \cdot e^{2x}}{(e^{2x} + 1)^2}\)
4Step 4: Simplifying the expression
After doing the multiplication and subtraction in the numerator, we have to simplify the expression: \(\frac{2e^{2x}(e^{2x} + 1) - 2e^{2x} \cdot e^{2x}}{(e^{2x} + 1)^2} = \frac{2e^{4x} + 2e^{2x} - 2e^{4x}}{(e^{2x} + 1)^2} = \frac{2e^{2x}}{(e^{2x} + 1)^2}\)
Key Concepts
Quotient Rule CalculusExponential Functions DerivativesApplying the Quotient Rule
Quotient Rule Calculus
The quotient rule is a technique in calculus for finding the derivative of a function that is the ratio of two differentiable functions. It states that the derivative of a quotient is equal to the derivative of the numerator times the denominator, minus the derivative of the denominator times the numerator, all over the denominator squared.
In formal terms, if we have a function represented as \(y = \frac{f(x)}{g(x)}\), where both \(f(x)\) and \(g(x)\) are differentiable, the derivative \(y'\) is given by:\[y' = \frac{f'(x) g(x) - f(x) g'(x)}{(g(x))^2}\]
To apply it effectively, remember to first identify the top and bottom functions and their derivatives separately before plugging into the rule. When simplifying your final answer, always check for opportunities to factor or cancel terms to achieve the most simplified form. Using the quotient rule can be tricky at first, but with practice, it becomes an invaluable tool in your calculus toolkit.
In formal terms, if we have a function represented as \(y = \frac{f(x)}{g(x)}\), where both \(f(x)\) and \(g(x)\) are differentiable, the derivative \(y'\) is given by:\[y' = \frac{f'(x) g(x) - f(x) g'(x)}{(g(x))^2}\]
To apply it effectively, remember to first identify the top and bottom functions and their derivatives separately before plugging into the rule. When simplifying your final answer, always check for opportunities to factor or cancel terms to achieve the most simplified form. Using the quotient rule can be tricky at first, but with practice, it becomes an invaluable tool in your calculus toolkit.
Exponential Functions Derivatives
Derivatives of exponential functions follow certain rules that make them more straightforward to calculate compared to other functions. If you have an exponential function of the form \(f(x) = e^{kx}\), where \(e\) is the base of the natural logarithm and \(k\) is a constant, then the derivative of that function is \(f'(x) = ke^{kx}\).
The key to understanding exponential derivatives is recognizing that the rate of change of an exponential function is directly proportional to the value of the function itself. This constant proportionality is what the constant multiplier \(k\) represents in the derivative.
For example, the derivative of \(f(x) = e^{2x}\) is \(f'(x) = 2e^{2x}\), which means the slope of the tangent at any point on the function's curve is twice the value of the function at that point. Being comfortable with this concept allows you to quickly differentiate a wide variety of exponential expressions.
The key to understanding exponential derivatives is recognizing that the rate of change of an exponential function is directly proportional to the value of the function itself. This constant proportionality is what the constant multiplier \(k\) represents in the derivative.
For example, the derivative of \(f(x) = e^{2x}\) is \(f'(x) = 2e^{2x}\), which means the slope of the tangent at any point on the function's curve is twice the value of the function at that point. Being comfortable with this concept allows you to quickly differentiate a wide variety of exponential expressions.
Applying the Quotient Rule
When applying the quotient rule, it's essential to work methodically and pay attention to detail. The correct application involves differentiating the top function (\(f(x)\)) and the bottom function (\(g(x)\)) separately, being careful with the signs when subtracting in the numerator, and ensuring that the denominator is correctly squared.
Let's take our example function \(y = \frac{e^{2x}}{e^{2x} + 1}\). The numerator (\(f(x)\)) is \(e^{2x}\), and the denominator (\(g(x)\)) is \(e^{2x} + 1\). Deriving both yields \(f'(x) = 2e^{2x}\) and \(g'(x) = 2e^{2x}\). Substituting these into the quotient rule gives us the derivative of the original function after simplification.
Nothing beats practice when it comes to mastering the application of the quotient rule. By applying it to various types of problems, including those with exponential functions, students will develop a deeper understanding and greater proficiency in the calculus of quotients.
Let's take our example function \(y = \frac{e^{2x}}{e^{2x} + 1}\). The numerator (\(f(x)\)) is \(e^{2x}\), and the denominator (\(g(x)\)) is \(e^{2x} + 1\). Deriving both yields \(f'(x) = 2e^{2x}\) and \(g'(x) = 2e^{2x}\). Substituting these into the quotient rule gives us the derivative of the original function after simplification.
Nothing beats practice when it comes to mastering the application of the quotient rule. By applying it to various types of problems, including those with exponential functions, students will develop a deeper understanding and greater proficiency in the calculus of quotients.
Other exercises in this chapter
Problem 50
Comparing Integration Problems In Exercises \(47-50\) , determine which of the integrals can be found using the basic integration formulas you have studied so f
View solution Problem 50
Finding a Derivative In Exercises \(37-58\) , find the derivative of the function. (Hint: In some exercises, you may find it helpful to apply logarithmic proper
View solution Problem 50
In Exercises 41–64, find the derivative of the function. $$ y=\ln \left[t\left(t^{2}+3\right)^{3}\right] $$
View solution Problem 50
Evaluating a Definite Integral In Exercises \(49-56\) , evaluate the definite integral. Use a graphing utility to verify your result. $$ \int_{-1}^{1} \frac{1}{
View solution