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\(s(y)=\frac{3 y}{y+5}\)

Short Answer

Expert verified
\(\frac{ds}{dy} = \frac{15}{(y+5)^2}\

Step by step solution

01

Apply Quotient Rule

The quotient rule in differential calculus states that if we have a function which is the ratio of two differentiable functions, say \(s(y) = \frac{u(v)}{v(w)}\), its derivative is given by \(\frac{s'(y)}{du/dy} = \frac{v(w) \cdot du/dv - u(v) \cdot dv/dw}{[v(w)]^2}\). We can apply this function to \(s(y)\), where our \(u(v) = 3y\) and \(v(w) = y+5\).
02

Calculate Derivatives

Next, we need to calculate the derivatives of \(u(v) = 3y\) and \(v(w) = y+5\). The derivative of \(u(v) = 3y\) with respect to \(y\) is \(3\), and the derivative of \(v(w) = y+5\) with respect to \(y\) is \(1\).
03

Substitute Back into Equation

Now substitute the derivatives calculated in Step 2 into the quotient rule equation. So our derivative \(\frac{ds}{dy} = \frac{(y+5) \cdot 3 - 3y \cdot 1}{(y+5)^2}\).
04

Simplify the Equation

Simplify the equation to arrive at the final derivative. So \(\frac{ds}{dy} = \frac{15}{(y+5)^2}\).

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Key Concepts

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

Quotient Rule
The Quotient Rule is a fundamental technique in differential calculus used when you need to find the derivative of a function that is the quotient of two other functions. Let's say we have a function in the form of \( f(x) = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are differentiable functions. According to the Quotient Rule, the derivative \( f'(x) \) is calculated as: \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^{2}}. \]

In simple terms, we take the derivative of the top function, multiply it by the bottom function, and from that, we subtract the product of the top function and the derivative of the bottom function. Finally, we divide the entire result by the square of the bottom function. This rule helps simplify the process of taking derivatives and is crucial for solving complex calculus problems.
Differential Calculus
Differential calculus is at the heart of understanding the rates at which quantities change. In simpler terms, it's the branch of mathematics that studies how functions change when their inputs change. The cornerstone of differential calculus is the concept of the derivative, which measures exactly how a function is changing at any given point. A derivative can be seen as the slope of the tangent line to a curve at a point or the instantaneous rate of change of the function.

Learning to calculate derivatives—the process known as differentiation—is essential in solving problems related to motion, optimization, and more. It's a powerful tool that gives us precise information about the behavior of functions and helps us predict and understand real-world phenomena.
Simplifying Expressions
Simplifying expressions is the process of making a mathematical expression easier to understand and work with. This can involve combining like terms, factoring, expanding products, and canceling common factors, or any other legal algebraic operation that leads to a simpler form. In the context of derivatives particularly, simplifying an expression helps in identifying and elucidating the behavior of the function more clearly.

Simplification may also involve minimizing the number of operations needed to calculate a value—this can be particularly useful in derivations where we are frequently working with complex expressions. By reducing an expression to its simplest form, the calculation of its derivative becomes more straightforward and less prone to errors, and it can make the interpretation of the derivative's behavior more intuitive.
Differentiable Functions
Differentiable functions are functions that have a derivative at all points of their domain. This means that the graph of a differentiable function has a tangent line at each interior point in its domain, and that the function's behavior can be analyzed and predicted locally by its derivative at those points. This property is central to all of differential calculus, as it guarantees that the techniques of differentiation can be applied.

In the realm of functions like \( s(y) = \frac{3y}{y+5} \), we first check to make sure that certain conditions – such as the denominator not being equal to zero – are met to verify that they are indeed differentiable in their domain. When you work with differentiable functions, utilizing rules like the Quotient Rule becomes a valid and incredibly useful way to find derivatives quickly and effectively.

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Most popular questions from this chapter

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The number of lumens (time rate of flow of light) \(L\) from a fluorescent lamp can be approximated by the model $$ L=-0.294 x^{2}+97.744 x-664.875, \quad 20 \leq x \leq 90 $$ where \(x\) is the wattage of the lamp. (a) Use a graphing utility to graph the function. (b) Use the graph from part (a) to estimate the wattage necessary to obtain 2000 lumens.

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