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Find the derivative of the function. \(y=\frac{-\sqrt{x^{2}+4}}{2 x^{2}}-\frac{1}{4} \ln \left(\frac{2+\sqrt{x^{2}+4}}{x}\right)\)

Short Answer

Expert verified
The derivative of the function \(y=\frac{-\sqrt{x^{2}+4}}{2 x^{2}}-\frac{1}{4} \ln\left(\frac{2+\sqrt{x^{2}+4}}{x}\right)\) is \(y' = \frac{x^2-\sqrt{x^2+4}*2x}{2x^4}-\frac{x^2-x\sqrt{\frac{2+\sqrt{x^{2}+4}}{x}}}{4*\frac{2+\sqrt{x^{2}+4}}{x}*x^2}\)

Step by step solution

01

Differentiate the rational part using the quotient rule

The derivative of the function \(y_1=\frac{-\sqrt{x^{2}+4}}{2 x^{2}}\) using the quotient rule \( (f / g)' = (gf' - fg') / g^2\) is \(y_1'=\frac{(2x^2)*(\frac{1}{2\sqrt{x^2+4}}*2x)-(-\sqrt{x^{2}+4)*(4x)}{(2x^2)^2}\)
02

Apply simplification

The derivative becomes \(y_1'=\frac{x^2-\sqrt{x^2+4}*2x}{2x^4}\)
03

Differentiate the logarithmic part

the derivative of the function \(y_2=-\frac{1}{4} \ln\left(\frac{2+\sqrt{x^{2}+4}}{x}\right)\) using the rule \( (a * ln u)' = a * (1 / u) * u' \) is \(y_2'= -\frac{1}{4}*\frac{1}{\frac{2+\sqrt{x^{2}+4}}{x}}*(\frac{\frac{1}{2\sqrt{\frac{2+\sqrt{x^{2}+4}}{x}}}*2x-\sqrt{\frac{2+\sqrt{x^{2}+4}}{x}}}{x^2}\)
04

Apply simplification

The derivative becomes \(y_2'=-\frac{x^2-x\sqrt{\frac{2+\sqrt{x^{2}+4}}{x}}}{4*\frac{2+\sqrt{x^{2}+4}}{x}*x^2}\)
05

Combine the derivatives

Since \(y=y_1+y_2\) the derivative of y is obtained by adding their derivatives. So, the derivative is \(y' = y_1' + y_2'=\frac{x^2-\sqrt{x^2+4}*2x}{2x^4}-\frac{x^2-x\sqrt{\frac{2+\sqrt{x^{2}+4}}{x}}}{4*\frac{2+\sqrt{x^{2}+4}}{x}*x^2}\)

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Key Concepts

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

Quotient Rule
When tackling the derivatives of rational functions, the quotient rule comes in handy. It's a procedure used when we want to differentiate a function that is the quotient of two other functions, typically expressed as \( f(x) / g(x) \). The quotient rule states that the derivative of this form \( (f / g)' \) is equal to \( (gf' - fg') / g^2 \), where \( f' \) and \( g' \) are the derivatives of \( f \) and \( g \) respectively.

The application of the quotient rule involves careful arithmetic and attention to detail. Starting with the identification of the numerator function \( f(x) \) and the denominator function \( g(x) \) and finding their derivatives. Once these components are determined, proceed by placing them into the quotient rule formula. For instance, with the function \( y_1=\frac{-\sqrt{x^{2}+4}}{2 x^{2}} \), the derivative is computed as \( y_1' \) by following these steps. Simplification of the result is essential to achieve the final derivative form, which comes down to handling algebraic expressions correctly.
Logarithmic Differentiation
Logarithmic differentiation is a technique that simplifies the differentiation of complex functions involving exponents or products by taking the natural logarithm (ln) of both sides. It's especially useful when differentiating a function that is the product of various terms that would be more daunting to differentiate individually.

In the exercise, logarithmic differentiation was applied to the function \( y_2=-\frac{1}{4} \ln\left(\frac{2+\sqrt{x^{2}+4}}{x}\right) \). To apply this technique, we first identify \( u \) as the inner function, which is the argument of the ln function. Then, using the rule that \( (a * \ln u)' = a * (1 / u) * u' \) the derivative becomes much more manageable. Subsequently, finding \( u' \), the derivative of \( u \), requires the use of other differentiation rules, like the chain rule or quotient rule, elegant examples of the interconnectivity within calculus methods. The simplification steps that follow ensure that the derivative is presented in its simplest form.
Chain Rule
The chain rule is crucial for differentiating compositions of functions, where one function is nested inside another. It states that if you have a composite function \( f(g(x)) \), its derivative \( f'(g(x))g'(x) \) can be found by multiplying the outer function's derivative, evaluated at the inner function \( f'(g(x)) \) by the derivative of the inner function \( g'(x) \).

This concept appears when differentiating complex expressions, such as the inside of a square root or inside a logarithm. Say, take the inner function \( g(x) \) and the outer function \( f(u) \) where \( u = g(x) \) - the chain rule helps here by differentiating \( f \) with respect to \( u \) and then multiplying by the derivative of \( g \) with respect to \( x \). This method is exemplified in the exercise when differentiating the argument of the logarithm function in \( y_2 \). The meticulous application of the chain rule is necessary to break down more complex derivatives into simpler terms and reach the correct solution.

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Most popular questions from this chapter

In Exercises 107-110, (a) use a graphing utility to find the derivative of the function at the given point, (b) find an equation of the tangent line to the graph of the function at the given point, and (c) use the utility to graph the function and its tangent line in the same viewing window. \(s(t)=\frac{(4-2 t) \sqrt{1+t}}{3},\left(0, \frac{4}{3}\right)\)

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Angle of Elevation A balloon rises at a rate of 3 meters per second from a point on the ground 30 meters from an observer. Find the rate of change of the angle of elevation of the balloon from the observer when the balloon is 30 meters above the ground.

The frequency \(F\) of a fire truck siren heard by a stationary observer is \(F=\frac{132,400}{331 \pm v}\) where \(\pm v\) represents the velocity of the accelerating fire truck in meters per second. Find the rate of change of \(F\) with respect to \(v\) when (a) the fire truck is approaching at a velocity of 30 meters per second (use \(-v)\) (b) the fire truck is moving away at a velocity of 30 meters per second (use \(+v\) ).

In Exercises \(81-88\), (a) find an equation of the tangent line to the graph of \(f\) at the indicated point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. \(\frac{\text { Function }}{y=4-x^{2}-\ln \left(\frac{1}{2} x+1\right)} \quad \frac{\text { Point }}{\left(0,4\right)}\)

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