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Find the first and second derivatives of the functions in Exercises 33-40. \(p=\frac{q^{2}+3}{(q-1)^{3}+(q+1)^{3}}\)

Short Answer

Expert verified
The first derivative is given by \(p'(q)\) and the second by \(p''(q)\); both require quotient rule and simplification.

Step by step solution

01

Differentiate the function using Quotient Rule

The given function is a quotient of two functions: numerator \(N(q) = q^2 + 3\) and denominator \(D(q) = (q-1)^3 + (q+1)^3\). Use the Quotient Rule for derivatives, which states: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).Calculate \(N'(q)\):- The derivative of \(q^2\) is \(2q\).- The derivative of the constant 3 is 0.- Thus, \(N'(q) = 2q\).Calculate \(D'(q)\):- The derivative of \((q-1)^3\) is \(3(q-1)^2\) using the chain rule.- The derivative of \((q+1)^3\) is \(3(q+1)^2\) using the chain rule.- Thus, \(D'(q) = 3(q-1)^2 + 3(q+1)^2\).Substitute back into the Quotient Rule formula:\[ p'(q) = \frac{(2q)((q-1)^3 + (q+1)^3) - (q^2 + 3)(3(q-1)^2 + 3(q+1)^2)}{((q-1)^3 + (q+1)^3)^2} \].
02

Simplify the First Derivative

Simplify the expression obtained from the derivative:1. Expand terms inside the numerator: - Calculate \((2q)((q-1)^3 + (q+1)^3)\). - Calculate \((q^2 + 3)(3(q-1)^2 + 3(q+1)^2)\).2. Combine and simplify any like terms.This step will give a more manageable expression for the first derivative \(p'(q)\). Simplifying may involve combining polynomials and gathering similar terms as you expand the expressions.
03

Calculate the Second Derivative

To find the second derivative \(p''(q)\), first confirm the simplified expression of \(p'(q)\). Then, differentiate \(p'(q)\) once more using standard differentiation rules:- You may need to use the quotient rule again if \(p'(q)\) is still a quotient.- Ensure derivatives of polynomials and chain rule are applied correctly.- Carefully simplify the result.\[ p''(q) = \frac{d}{dq}\left(p'(q)\right) \].
04

Substitute and Simplify Second Derivative

Substitute the derivatives found in the previous steps back into the earlier formulas.Simplify the second derivative expression to obtain \(p''(q)\) in its simplest polynomial or fractional form. This step may involve collecting like terms, simplifying fractions, or expanding multiplied constants.

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Key Concepts

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

Quotient Rule
When you want to find the derivative of a function that is the quotient of two other functions, you use the Quotient Rule. This is especially helpful for functions that look like fractions. The rule itself states that for two functions, say \( u \) and \( v \), the derivative of their quotient \( \frac{u}{v} \) is:\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]Here, \( u' \) and \( v' \) represent the derivatives of the numerator and the denominator, respectively. The "prime" notation (\( ' \)) is a shorthand for the derivative. Using this rule helps you systematically find the derivatives of complex fractional expressions in calculus settings.
The Quotient Rule avoids mistakes by ensuring each part of the fraction contributes correctly to the final derivative, accounting for the structure of how the functions are divided.
First Derivative
Finding the first derivative, denoted as \( p'(q) \) for a function \( p(q) \), allows you to understand the rate of change of the function at any given point. In our example, to find the first derivative of \( p(q) = \frac{q^{2}+3}{(q-1)^{3}+(q+1)^{3}} \), we apply the Quotient Rule.
  • First, find the derivatives of the numerator \( N(q) = q^2 + 3 \), which is simply \( N'(q) = 2q \).
  • Then, tackle the denominator \( D(q) = (q-1)^3 + (q+1)^3 \). By using the Chain Rule, as the denominator consists of composite functions, we find \( D'(q) = 3(q-1)^2 + 3(q+1)^2 \).
Substituting these derivatives into the Quotient Rule formula gives the expression for \( p'(q) \), which can then be simplified.
Simplifying the first derivative involves expanding and combining terms to make your solution clearer and easier to work with for further calculations.
Second Derivative
The second derivative, denoted as \( p''(q) \), is the derivative of the first derivative and gives us information about the curvature or the concavity of the original function. Finding the second derivative involves differentiating \( p'(q) \) once more.
  • Firstly, simplify the expression obtained in the first derivative step, getting rid of unnecessary complexities.
  • Then, check if the function is still a quotient. If so, use the Quotient Rule again; otherwise, apply standard differentiation rules.
  • It may involve further applications of the Chain Rule for composite functions within \( p'(q) \).
Once differentiated, simplifying the result will give you \( p''(q) \) in its simplest form, revealing more about the intervals of increasing or decreasing rates of change in \( p(q) \). Keep in mind, the second derivative provides insight into the behavior of the function regarding its acceleration and inflection points.
Chain Rule
The Chain Rule is an essential principle in calculus used to differentiate the composition of functions. It helps when you have a function inside another function; for example, in expressions like \((q-1)^3\) or \((q+1)^3\).The Chain Rule states that if you have a composite function \( f(g(x)) \), its derivative is:\[ \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) \]This means you take the derivative of the outer function \( f \), and multiply by the derivative of the inner function \( g \).
In our context, when we differentiated \( (q-1)^3 \) or \( (q+1)^3 \), we:- Applied the power rule to get \( 3(q-1)^2 \),- Multiplied by the derivative \( 1 \) of \( (q-1) \) or \((q+1)\). This step is crucial as it respects the nested structure of the function.
The Chain Rule is vital for maintaining the integrity of derivatives when functions are embedded within each other.

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Most popular questions from this chapter

In Exercises \(35-40,\) write a differential formula that estimates the given change in volume or surface area. $$ \begin{array}{l}{\text { The change in the lateral surface area } S=\pi r \sqrt{r^{2}+h^{2}} \text { of a right }} \\ {\text { circular cone when the radius changes from } r_{0} \text { to } r_{0}+d r \text { and the }} \\\ {\text { height does not change }}\end{array} $$

a. Find the tangent to the curve \(y=2 \tan (\pi x / 4)\) at \(x=1\) b. Slopes on a tangent curve What is the smallest value the slope of the curve can ever have on the interval \(\quad-2

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The linearization is the best linear approximation Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=\) \(m(x-a)+c\) is a linear function in which \(m\) and \(c\) are constants. If the error \(E(x)=f(x)-g(x)\) were small enough near \(x=a\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions $$ \begin{array}{l}{\text { 1. } E(a)=0} \\ {\text { 2. } \lim _{x \rightarrow a} \frac{E(x)}{x-a}=0}\end{array} $$ then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization \(L(x)\) gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \(x-a\) .

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