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\(37-48\) Use logarithmic differentiation to find the derivative of the function. $$y=\sqrt{x} e^{x^{2}}\left(x^{2}+1\right)^{10}$$

Short Answer

Expert verified
The derivative is \( \sqrt{x} e^{x^{2}} (x^{2}+1)^{10} \left( \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right) \)."

Step by step solution

01

Express the Function in Logarithmic Form

Begin by taking the natural logarithm of both sides of the equation to simplify differentiation. Given:\[ y = \sqrt{x} e^{x^{2}} (x^{2}+1)^{10} \] Taking logs:\[ \ln y = \ln (\sqrt{x}) + \ln (e^{x^{2}}) + \ln ((x^{2}+1)^{10}) \] This becomes:\[ \ln y = \frac{1}{2} \ln x + x^{2} + 10 \ln (x^{2}+1) \]
02

Apply Differentiation Using Logarithmic Properties

Differentiate both sides with respect to \( x \). Remember that the derivative of \( \ln y \) with respect to \( x \) is \( \frac{1}{y} \frac{dy}{dx} \).\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{2x} + 2x + 10 \cdot \frac{2x}{x^2+1} \]
03

Solve for \( \frac{dy}{dx} \)

Now, solve for \( \frac{dy}{dx} \) by multiplying both sides by \( y \).\[ \frac{dy}{dx} = y \left( \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right) \] Substituting back for \( y \):\[ \frac{dy}{dx} = \sqrt{x} e^{x^{2}} (x^{2}+1)^{10} \left( \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right) \]

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Key Concepts

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

Derivative of a Function
Derivatives represent the rate of change of a function. In calculus, finding a derivative is essential for understanding how a function behaves. When differentiating more complex functions, like products or compositions of simple functions, several rules and techniques can be employed.
One such technique is logarithmic differentiation, useful when the function involves several multiplicative components. It allows the process of differentiation to be simplified, especially with functions of the form \( f(x) = u(x)v(x)w(x) \).
For a given function \( y = f(x) \), the derivative is denoted by \( \frac{dy}{dx} \), representing the rate at which \( y \) changes with respect to \( x \). By differentiating the equation step-by-step, we can break the problem into smaller, manageable parts. Here, transforming the initial function into a logarithmic form simplifies finding the derivative.
Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is a logarithm with base \( e \), where \( e \approx 2.71828 \) is the Euler's number. This mathematical function is crucial in solving problems involving growth and decay, and appears frequently in calculus due to its nice properties during differentiation and integration.
One of its key properties is that the derivative of \( \ln(x) \) is \( \frac{1}{x} \), which significantly simplifies differentiation.
In the case of logarithmic differentiation, we often take the logarithm of both sides of an equation where the function contains terms that are products, quotients, or powers. This converts multiplicative relationships into additive ones, making it easier to apply the basic derivative rules.
Chain Rule
The chain rule is a fundamental theorem for computing the derivative of a composite function. If you have two functions, say \( g(x) \) and \( f(u) \), where \( u = g(x) \), then the composite function is \( f(g(x)) \).
The chain rule states that the derivative of this composite is found by multiplying the derivative of the outer function by the derivative of the inner function:
\[ \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \]
In essence, the chain rule is crucial when dealing with nested functions or when a function is expressed as the composition of two or more functions. It allows us to "peel back" the layers of a function to fully understand how each part contributes to the whole derivative.
In the context of our solution, after taking the natural logarithm, we differentiated using the chain rule to handle each component of the logarithm separately.
Exponential Function
Exponential functions have the general form \( a^x \) or \( e^x \), where \( e \) is Euler's number. They are pivotal in mathematical modeling of real-world phenomena, such as population growth, radioactive decay, and interest calculations.
A distinguishing feature of the exponential function \( e^x \) is that it is its own derivative: this means the slope of \( e^x \) at any point is equal to its value at that point.
In our exercise, the function \( e^{x^2} \) appears. To differentiate such expressions, we often employ the chain rule, considering \( e^{x^2} \) as the outer function and \( x^2 \) as the inner function. By understanding the properties of exponential functions and utilizing the chain rule, you can tackle these derivatives efficiently.

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Most popular questions from this chapter

If \(R\) denotes the reaction of the body to some stimulus of strength \(x,\) the sensitivity \(S\) is defined to be the rate of change of the reaction with respect to \(x\) . A particular example is that when the brightness \(x\) of a light source is increased, the eye reacts by decreasing the area \(R\) of the pupil. The experimental formula $$R=\frac{40+24 x^{0.4}}{1+4 x^{0.4}}$$ has been used to model the dependence of \(\mathrm{R}\) on \(\mathrm{x}\) when \(\mathrm{R}\) is measured in square millimeters and \(\mathrm{x}\) is measured in appropriate units of brightness. (a) Find the sensitivity. (b) Illustrate part (a) by graphing both \(\mathrm{R}\) at low levels of of \(\mathrm{x} .\) Comment on the values of \(\mathrm{R}\) and \(\mathrm{S}\) at low levels of brightness. Is this what you would expect?

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