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Chapter 3: Problem 12

Find the derivative of the given function. $$ G(x)=\frac{4 x+6}{\sqrt{x^{2}+3 x+4}} $$

Short Answer

Expert verified
\( G'(x) = \frac{3x + 7}{(x^2 + 3x + 4)^{3/2}} \)

Step by step solution

01

- Identify the function components

The given function is a quotient of two functions: the numerator is \(4x + 6\) and the denominator is \(\sqrt{x^2 + 3x + 4}\).
02

- Apply the Quotient Rule

The quotient rule states that if you have a function \(\frac{u(x)}{v(x)}\), then the derivative is given by \(\frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}\). Here, \(u(x) = 4x + 6\) and \(v(x) = \sqrt{x^2 + 3x + 4}\).
03

- Find the derivative of the numerator

Differentiate \(u(x) = 4x + 6\). The derivative is \(u'(x) = 4\).
04

- Find the derivative of the denominator

First, rewrite the denominator \(v(x) = (x^2 + 3x + 4)^{1/2}\). Then use the chain rule to differentiate it. The chain rule gives \(v'(x) = \frac{1}{2}(x^2 + 3x + 4)^{-1/2} (2x + 3) = \frac{2x + 3}{2\sqrt{x^2 + 3x + 4}}\).
05

- Apply the Quotient Rule Formula

Now that you have \(u'(x) = 4\), \(u(x) = 4x + 6\), \(v(x) = \sqrt{x^2 + 3x + 4}\), and \(v'(x) = \frac{2x + 3}{2\sqrt{x^2 + 3x + 4}}\), plug these into the quotient rule formula: \[ G'(x) = \frac{(4)(\sqrt{x^2 + 3x + 4}) - (4x + 6)\left(\frac{2x + 3}{2\sqrt{x^2 + 3x + 4}}\right)}{(\sqrt{x^2 + 3x + 4})^2} \]. Simplify the numerator: \[ 4\sqrt{x^2 + 3x + 4} - (4x + 6)\frac{2x + 3}{2\sqrt{x^2 + 3x + 4}} \]. Multiply to combine terms over a common denominator: \[ 4\sqrt{x^2 + 3x + 4} - \frac{(4x + 6)(2x + 3)}{2\sqrt{x^2 + 3x + 4}} \].
06

- Simplify the expression

Combine like terms in the numerator: \[ \frac{8(x^2 + 3x + 4) - (4x + 6)(2x + 3)}{2\sqrt{x^2 + 3x + 4}(x^2 + 3x + 4)} \]. Further simplify the numerator: \[ 8(x^2 + 3x + 4) - (8x^2 + 18x + 18) = \frac{8x^2 + 24x + 32 - 8x^2 - 18x - 18}{2(x^2 + 3x + 4)^{3/2}} \]. Combine like terms: \[ \frac{6x + 14}{2(x^2 + 3x + 4)^{3/2}} = \frac{3x + 7}{(x^2 + 3x + 4)^{3/2}} \].

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Key Concepts

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

Quotient Rule
When dealing with derivatives of functions that are represented as fractions, the Quotient Rule is a powerful tool. The rule states that if you have a function in the form \(\frac{u(x)}{v(x)}\), then the derivative of this function is found using the formula: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]. Breaking it down:
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Most popular questions from this chapter

Differentiate the given function by applying the theorems of this section. $$ f(x)=\left(2 x^{4}-1\right)\left(5 x^{3}+6 x\right) $$

In Exercises 1 through 14, do each of the following: (a) Draw a sketch of the graph of the function; (b) determine if \(f\) is continuous at \(x_{1} ;\) (c) find \(f^{\prime}-\left(x_{1}\right)\) and \(f_{+}^{\prime}\left(x_{1}\right)\) if they exist; (d) determine if \(f\) is differentiable at \(x_{1}\). $$ \begin{gathered} f(x)=\left\\{\begin{aligned} x+2 & \text { if } x \leq-4 \\ -x-6 & \text { if } x>-4 \end{aligned}\right. \\ x_{1}=-4 \end{gathered} $$

Find an equation of the tangent line to the curve \(y=\sqrt{x^{2}+9}\) at the point \((4,5)\).

In Exercises 24 through 27 , a particle is moving along a straight line according to the given equation of motion, where \(s\) \(\mathrm{ft}\) is the directed distance of the particle from the origin at \(t \mathrm{sec}\). Find the time when the instantaneous acceleration is zero, and then find the directed distance of the particle from the origin and the instantaneous velocity at this instant. $$ s=2 t^{3}-6 t^{2}+3 t-4, t \geq 0 $$

Find the derivative of the given function. $$ f(x)=\left(4 x^{2}+7\right)^{2}\left(2 x^{3}+1\right)^{4} $$
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