Problem 12 - Calculus an Applied Approach

Question

Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$\begin{array}{ll}{\text { Function }} & {\text { Point }} \\ {f(x)=\frac{3 x}{x^{2}+1}} & {\left(-1,-\frac{3}{5}\right)} \end{array}$$

Step-by-Step Solution

Answer

The value of the derivative of the function at the point \(-1, -\frac{3}{2}\) is 0. The differentiation rule used is the quotient rule.

Step 1: Differentiate the function

To find the derivative of the function \(f(x) = \frac{3x}{{x^2+1}}\), remember that the quotient rule states that the derivative of \(\frac{u}{v}\) is \(\frac{vu'-uv'}{{v^2}}\), where \(u'\) and \(v'\) denote the derivatives of \(u\) and \(v\) respectively. For \(f(x)\), \(u = 3x\) and \(v = x^2 + 1\). So start by finding \(u'\) and \(v'\). \(u' = 3\) and \(v' = 2x\). By applying the quotient rule, \(f'(x) = \frac{v*u' - u*v'}{{v^2}} = \frac{(x^2+1)*3 - 3x*2x}{{(x^2+1)^2}} = \frac{3x^2+3 -6x^2}{{(x^2+1)^2}} = \frac{-3x^2 + 3}{{(x^2+1)^2}}.

Step 2: Evaluate the derivative at the given point

Now, put \(x = -1\) into \(f'(x)\) to get the derivative of the function at the given point \(-1, -\frac{3}{2}\). So, \(f'(-1) = \frac{-3*(-1)^2 + 3}{{((-1)^2+1)^2}} = \frac{0}{4} = 0.

Step 3: State the differentiation rule used

The differentiation rule used in this exercise is the quotient rule for differentiation.

Key Concepts

Quotient Rule

Differentiation can often be tackled using several key rules, and one such rule is the Quotient Rule. The Quotient Rule is a method used in calculus to find the derivative of a function that is the division of two differentiable functions. If you have a function in the form \(f(x) = \frac{u(x)}{v(x)}\), the Quotient Rule states that the derivative \(f'(x)\) is:

• Take the derivative of the top function \(u(x)\), denoted as \(u'(x)\).
• Take the derivative of the bottom function \(v(x)\), denoted as \(v'(x)\).
• Plug these into the formula: \(f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{v(x)^2}\).

This rule efficiently handles the differentiation of ratios, saving you from the cumbersome work of expanding the function entirely. In our given exercise \(f(x) = \frac{3x}{x^2 + 1}\), we set \(u(x) = 3x\) and \(v(x) = x^2 + 1\). Utilizing the Quotient Rule, we first find \(u'(x) = 3\) and \(v'(x) = 2x\) and substitute them into the formula to find the derivative.

Derivative Evaluation

Once you've effectively differentiated a function using the appropriate rule, the next task is often to evaluate this derivative at a specific point. This means plugging the x-value of that point into the derivative function, which gives us the slope of the tangent line to the curve at that particular x-value.

For the given problem, after applying the Quotient Rule, we found that the derivative of \(f(x) = \frac{3x}{x^2 + 1}\) is \(f'(x) = \frac{-3x^2 + 3}{(x^2 + 1)^2}\). Now, to find the derivative at \(x = -1\), simply substitute \(-1\) into the derivative expression. This substitution simplifies to:

• \(f'(-1) = \frac{-3(-1)^2 + 3}{((-1)^2 + 1)^2} = \frac{0}{4} = 0\)

This calculation tells us that the slope of the tangent line to the graph at the point \((-1, -\frac{3}{5})\) is 0, indicating a horizontal tangent.

Calculus Problem

Solving calculus problems often requires choosing and applying the correct differentiation rules and carefully evaluating the results. The problem in our exercise highlights an important aspect of calculus: the need to analyze and simplify expressions properly using rules like the Quotient Rule.

When presented with calculus problems, it is crucial to:

• Identify the form of the function (product, quotient, chain, etc.).
• Select the appropriate differentiation rule.
• Execute the rule step-by-step, as with finding \(u'\) and \(v'\).
• Finally, evaluate the derivative at any given point, if necessary.

By systematically applying these steps, you gain not only the solution but a deeper understanding of the underlying mechanics of calculus. This kind of practice lets you tackle a wide range of differentiation tasks, building post foundational skills that are essential across many fields of study and work.