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Finding a Derivative IIn Exercises \(9-34,\) find the derivative of the function. \(y=\frac{x}{\sqrt{x^{4}+4}}\)

Short Answer

Expert verified
The derivative of the function \(y\) is \(\frac{-x^4}{x^4+4}\)

Step by step solution

01

The Quotient Rule

The quotient rule of differentiation states that the derivative of \(\frac{u}{v}\) is \(\frac{v(u') - u(v')}{v^2}\) where \(u'\) and \(v'\) are the derivatives of \(u\) and \(v\) respectively.
02

Identify u and v

Here \(u=x\) and \(v=\sqrt{x^{4}+4}\). Hence, \(u'=1\) and for finding \(v'\), the chain rule needs to be applied
03

Calculate Derivative using Chain Rule

The Chain Rule states that if you compose two functions to create a third function, lets say \(h(x) = f(g(x))\), then the derivative of \(h(x)\) with respect to \(x\) is equal to the derivative of \(f\) with respect to \(g(x)\) multiplied by the derivative of \(g(x)\) with respect to \(x\). Hence, \(v' = \frac{1}{2\sqrt{x^4+4}} * 4x^{3}\)
04

Substitute values in Quotient Rule

Substitute the values \(u\), \(v\), \(u'\), \(v'\) in the quotient rule formula to get \(\frac{1*x^4 - 2*x^4}{(x^4+4)} = \frac{-x^4}{x^4+4}\)

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Key Concepts

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

Derivative
In calculus, a derivative represents how a function changes as its input changes. It provides the rate at which a quantity varies with respect to another quantity. In simpler terms, the derivative is like the slope of a line that shows how steep or flat it is at any point.
To find the derivative of a function, you can use various techniques depending on the form of the function. For instance, when dealing with a fraction, like in our exercise, the quotient rule becomes very useful. Knowing the rules of derivatives, like the power rule, product rule, and quotient rule, helps simplify and accurately determine the derivatives.
Chain Rule
The chain rule is an essential tool in calculus, especially when dealing with composite functions. Composite functions are functions nested within each other, such as \(h(x) = f(g(x))\).
To find the derivative of such a composite function, the chain rule instructs us to take the derivative of the outer function \(f\) with respect to the inner function \(g(x)\), and multiply it by the derivative of \(g(x)\) with respect to \(x\).
In the original exercise, we used the chain rule to find the derivative of \(v = \sqrt{x^4 + 4}\). This expression was simplified to \(v' = \frac{1}{2\sqrt{x^4 + 4}} \times 4x^3\), demonstrating how the chain rule helped us tackle the inner component \(x^4 + 4\).
Calculus
Calculus is a branch of mathematics that studies continuous change. It consists of two main concepts: differentiation, which deals with finding rates of change, and integration, which involves finding accumulations. These are used to investigate changes within a system.
In many real-world applications, calculus helps us analyze changes that occur and predict future behavior. Scientists and engineers rely on calculus for modeling and solving problems. In this exercise, we're focusing on differentiation by applying rules like the quotient rule to find how a function changes.
Differentiation
Differentiation is the process of finding a derivative. It's a fundamental concept in calculus used to understand how functions behave. Differentiation helps us find the slope at any point on a curve, which illustrates how the function's output changes with respect to its input.
In the given exercise, we used differentiation to figure out the rate of change of the function \(y = \frac{x}{\sqrt{x^4 + 4}}\). By applying differentiation rules like the quotient and chain rules, we systematically broke down the function step by step to find the derivative, \(\frac{-x^4}{x^4 + 4}\). These tools in differentiation provide insights into understanding complex functions.

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Most popular questions from this chapter

Finding an Equation of a Exercises \(71-78\) , (a) find an equation of the tangent line to the graph of the function at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the tangent feature of a graphing utility to confirm your results. $$y=\cos 3 x,\left(\frac{\pi}{4},-\frac{\sqrt{2}}{2}\right)$$

Determining Differentiability In Exercises \(85-88\) , find the derivatives from the left and from the right at \(x=1\) (if they exist). Is the function differentiable at \(x=1 ?\) $$f(x)=\left\\{\begin{array}{ll}{(x-1)^{3},} & {x \leq 1} \\ {(x-1)^{2},} & {x>1}\end{array}\right.$$

pumped into the trough. Moving Ladder A ladder 25 feet long is leaning against the wall of a house (see figure). The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) How fast is the top of the ladder moving down the wall when its base is 7 feet, 15 feet, and 24 feet from the wall? (b) Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall. (c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 7 feet from the wall.

Finding an Equation of a Exercises \(71-78\) , (a) find an equation of the tangent line to the graph of the function at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the tangent feature of a graphing utility to confirm your results. $$y=\left(4 x^{3}+3\right)^{2},(-1,1)$$

FOR FURTHER INFORMATION For more information on the mathematics of moving ladders, see the article "The Falling Ladder Paradox" by Paul Scholten and Andrew Simoson in The College Mathematics Journal. To view this article, go to MathArticles.com. Construction A construction worker pulls a five-meter plank up the side of a building under construction by means of a rope tied to one end of the plank (see figure). Assume the opposite end of the plank follows a path perpendicular to the wall of the building and the worker pulls the rope at a rate of 0.15 meter per second. How fast is the end of the plank sliding along the ground when it is 2.5 meters from the wall of the building?

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