Problem 52
Question
In Problems 47-58, express the indicated derivative in terms of the function \(F(x)\). Assume that \(F\) is differentiable. $$ \frac{d}{d y}\left(y^{2}+\frac{1}{F\left(y^{2}\right)}\right) $$
Step-by-Step Solution
Verified Answer
\( 2y - \frac{2yF'(y^2)}{F(y^2)^2} \)
1Step 1: Identify the function structure
The expression given is \( y^2 + \frac{1}{F(y^2)} \). We need to determine the derivative with respect to \( y \). We will use the chain rule and the power rule.
2Step 2: Differentiate \( y^2 \)
The first term \( y^2 \) can be directly differentiated: \( \frac{d}{dy}(y^2) = 2y \).
3Step 3: Differentiate \( \frac{1}{F(y^2)} \) using the chain rule
Consider \( u = F(y^2) \), then \( \frac{1}{u} \) has to be differentiated. The derivative \( \frac{d}{dy} \left( \frac{1}{u} \right) \) is \( -\frac{1}{u^2} \cdot \frac{du}{dy} \).
4Step 4: Differentiate \( F(y^2) \) using the chain rule
To find \( \frac{du}{dy} = \frac{d}{dy} F(y^2) \), we apply the chain rule: let \( v = y^2 \), so \( \frac{d}{dv} F(v) = F'(v) \). Then multiply by \( \frac{dv}{dy} = 2y \), giving \( F'(y^2) \cdot 2y \).
5Step 5: Combine results for \( \frac{d}{dy} \left( \frac{1}{F(y^2)} \right) \)
Substitute \( u = F(y^2) \) and \( \frac{du}{dy} = F'(y^2) \cdot 2y \) into the chain rule equation to get: \( -\frac{1}{F(y^2)^2} \cdot F'(y^2) \cdot 2y \). Simplify this to \( -\frac{2yF'(y^2)}{F(y^2)^2} \).
6Step 6: Combine derivatives
Add the derivatives of both terms together to get the result: \( \frac{d}{dy} \left(y^2 + \frac{1}{F(y^2)}\right) = 2y - \frac{2yF'(y^2)}{F(y^2)^2} \).
Key Concepts
Chain RulePower RuleDifferentiable FunctionsCalculus Problem Solving
Chain Rule
The Chain Rule is an essential concept in calculus for differentiating compositions of functions. When a function is nested within another function, the chain rule helps us find the derivative of the outer function with respect to the inner one, and then multiply it by the derivative of the inner function.
In our example, we use the chain rule to differentiate the expression \( \frac{1}{F(y^2)} \). Here, think of \( F(y^2) \) as a function inside an outer function \( \frac{1}{x} \), where \( x = F(y^2) \).
This tells us to first differentiate \( \frac{1}{x} \) which is \( -\frac{1}{x^2} \), and then multiply by the derivative of \( F(y^2) \), using the chain rule. This derivative is \( F'(y^2) \cdot 2y \), as derived from differentiating \( u = F(y^2) \).
In our example, we use the chain rule to differentiate the expression \( \frac{1}{F(y^2)} \). Here, think of \( F(y^2) \) as a function inside an outer function \( \frac{1}{x} \), where \( x = F(y^2) \).
This tells us to first differentiate \( \frac{1}{x} \) which is \( -\frac{1}{x^2} \), and then multiply by the derivative of \( F(y^2) \), using the chain rule. This derivative is \( F'(y^2) \cdot 2y \), as derived from differentiating \( u = F(y^2) \).
Power Rule
The Power Rule is one of the simplest and most commonly used rules in calculus for finding derivatives. It helps us to differentiate polynomials efficiently. For a function like \( y^n \) where \( n \) is any real number, the power rule states that the derivative is \( n \cdot y^{n-1} \).
In our given problem, we apply the power rule to the term \( y^2 \).
By using the rule, we bring down the exponent as a coefficient and reduce the exponent by one, giving us \( \frac{d}{dy}(y^2) = 2y \). This step is straightforward, emphasising that recognizing basic derivatives is crucial for solving calculus problems swiftly.
In our given problem, we apply the power rule to the term \( y^2 \).
By using the rule, we bring down the exponent as a coefficient and reduce the exponent by one, giving us \( \frac{d}{dy}(y^2) = 2y \). This step is straightforward, emphasising that recognizing basic derivatives is crucial for solving calculus problems swiftly.
Differentiable Functions
Differentiable functions are those that have a derivative at every point in their domain. This means we can find the slope of the tangent line at any point on the graph of the function.
In this exercise, we assume that \( F \) is differentiable. This implies that \( F \) is smooth and continuous, allowing us to apply the derivative rules conveniently. Since \( F(y^2) \) is involved, it is crucial that \( F \) can be differentiated to maintain the integrity of our solution.
When working with differentiable functions, especially in compositions like \( F(y^2) \), it's essential to check that the function satisfies the differentiability condition across the interval of interest.
In this exercise, we assume that \( F \) is differentiable. This implies that \( F \) is smooth and continuous, allowing us to apply the derivative rules conveniently. Since \( F(y^2) \) is involved, it is crucial that \( F \) can be differentiated to maintain the integrity of our solution.
When working with differentiable functions, especially in compositions like \( F(y^2) \), it's essential to check that the function satisfies the differentiability condition across the interval of interest.
Calculus Problem Solving
Solving calculus problems often involves breaking down complex expressions into simpler parts. Our example demonstrates this by separating the terms \( y^2 \) and \( \frac{1}{F(y^2)} \) to handle each piece with appropriate rules.
By applying the chain rule and power rule separately, we handle the individual tasks of differentiation more effectively.
Here's how you can approach calculus problems:
By applying the chain rule and power rule separately, we handle the individual tasks of differentiation more effectively.
Here's how you can approach calculus problems:
- Identify the structure of the function you need to differentiate.
- Apply known rules, like the chain rule or power rule, to each segment.
- Combine each differentiated part to form the complete derivative.
Other exercises in this chapter
Problem 52
\(y=x^{3}-3 x^{2}\)
View solution Problem 52
Find all points on the graph of \(y=\frac{1}{3} x^{3}+x^{2}-x\) where the tangent line has slope 1 .
View solution Problem 53
Find \(d y / d x\) by logarithmic differentiation \(y=\frac{x+11}{\sqrt{x^{3}-4}}\)
View solution Problem 53
Find all points on the graph of \(y=100 / x^{5}\) where the tangent line is perpendicular to the line \(y=x\).
View solution