Question

Express the indicated derivative in terms of the function \(F(x) .\) Assume that \(F\) is differentiable. $$ \frac{d}{d y}\left(y^{2}+\frac{1}{F\left(y^{2}\right)}\right) $$

Short answer

The derivative is \(2y - \frac{2yF'(y^2)}{(F(y^2))^2}\).

Step-by-step solution

Differentiate the First Term

The first term in the expression is \(y^2\). The derivative of \(y^2\) with respect to \(y\) is \(2y\).

Apply the Chain Rule to the Second Term

The second term is \(\frac{1}{F(y^2)}\). Rewrite this as \((F(y^2))^{-1}\). Use the power rule combined with the chain rule for differentiation, and consider \(F(y^2)\) as \(u\), so \(\frac{d}{dy}(u^{-1}) = -u^{-2}\cdot\frac{du}{dy}\).

Differentiate \(u = F(y^2)\) using Chain Rule

Identify \(u = F(y^2)\). To find \(\frac{du}{dy}\), apply the chain rule: \(\frac{du}{dy} = F'(y^2) \cdot \frac{d}{dy}(y^2) = F'(y^2) \cdot 2y\).

Plug \(\frac{du}{dy}\) Back into the Derivative from Step 2

Substitute \(\frac{du}{dy} = 2yF'(y^2)\) into the expression from Step 2: \(- (F(y^2))^{-2} \cdot 2yF'(y^2)\). This simplifies to \(- \frac{2yF'(y^2)}{(F(y^2))^2}\).

Combine the Results

Add the results from Steps 1 and 4: \(2y - \frac{2yF'(y^2)}{(F(y^2))^2}\). This is the derivative of the entire expression with respect to \(y\).

Key concepts

Calculus Differentiation

Calculus differentiation is a fundamental concept that deals with finding the rate at which a function changes as its input changes. Imagine you have a curve on a graph, and differentiation helps you determine the slope of this curve at any given point. It's like finding out how fast you're traveling at an exact moment.

There are several basic rules in differentiation that help simplify the process:

• **Power Rule**: This rule is used whenever you have to differentiate a function that is a power of a variable, like in our exercise with the derivative of the term \(y^2\). The power rule states that the derivative of \(x^n\) is \(nx^{n-1}\).
• **Constant Rule**: Any constant value, when differentiated, simply becomes zero.
• **Sum Rule**: If you have two functions added together, the derivative of the sum is the sum of their derivatives.

Having these rules in your toolkit makes handling differentiation tasks much easier. They allow you to break down complex functions into simpler parts and handle them one step at a time.

Derivatives

In the context of calculus, a derivative represents the rate of change of a function. It's like peeking into the function's behavior at a tiny, specific point, to see how it evolves. The derivative is often denoted by \(f'(x)\), \(\frac{dy}{dx}\) or similar expressions, indicating it’s about change with respect to the variable \(x\).

Understanding derivatives involves several key ideas:

• **Slope of a Tangent Line**: Derivatives give you the slope of the tangent line to the curve of a function at a given point. This slope tells you how steep the curve is, providing insights into the increasing or decreasing nature of the function.
• **Instantaneous Rate of Change**: Think of it as zooming into a very tiny spot on the curve to see how quickly or slowly the function value changes.
• **Applications in Real Life**: Derivatives can be used to find velocity in physics when given a position over time, or to determine how the concentration of a substance changes over time in chemistry.

In our original exercise, the derivative was expressed in terms of the function \(F(x)\), demonstrating how derivatives can be applied to functions of other functions, using rules like the chain rule.

Chain Rule Application

The chain rule is a crucial technique in calculus for differentiating compositions of functions. It's like peeling away the layers of an onion, addressing each layer one at a time. If a function is nested inside another, the chain rule helps us find the derivative of the entire composite function. For example, if you have a function \(g(x) = f(h(x))\), the chain rule says that the derivative \(g'(x)= f'(h(x)) \cdot h'(x)\).

To apply the chain rule effectively, you should:

• **Identify the Layers**: Find which function is inside of which, like seeing \(u\) in \((u)^{-1}\) where \(u = F(y^2)\).
• **Differentiate the Outer Function**: Take the derivative of the outer function first, as done in step 2 of the solution.
• **Differentiate the Inner Function**: Then differentiate the inner layer, considering their respective variable changes.
• **Multiply the Derivatives**: Finally, multiply the derivatives of the outer and inner functions.

In our original solution, the chain rule was essential for differentiating the composite function \((F(y^2))^{-1}\), systematically breaking down into manageable parts.

Function Differentiation

Function differentiation involves applying rules such as the power, product, and chain rules to find the derivative of functions. When dealing with more intricate expressions, such as compositions of functions, differentiation requires methodical application of these rules.

Some important points include:

• **Composite Functions**: Involving multiple layers of functions, such as \(F(y^2)\), where the composition requires careful application of the chain rule.
• **Breaking Down Components**: Treat each function layer separately, differentiate, and then recombine the results, as demonstrated in the solution to find the total derivative.
• **Simplification**: Once you find the derivatives, simplification is often necessary to present the expression in a more concise form.

In the provided exercise, the differentiation of the function \(\frac{d}{dy}(y^2 + \frac{1}{F(y^2)})\) showcases how these techniques come together to compute the overall derivative efficiently. Practice makes these methods easy to apply, especially by working through exercises like this one.