Question

Finding an Equation of a Tangent Line In Exercises \(73-80,\) (a) find an equation of the tangent line to the graph of \(f\) at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of the graphing utility to confirm your results. $$ f(x)=\frac{1}{3} x \sqrt{x^{2}+5},(2,2) $$

Short answer

The equation of the tangent line to the function \(f(x)\) at the point (2,2) is \(y - 2 = 8/3(x - 2)\)

Step-by-step solution

Derivation of the function

To find the derivative of the function \(f(x) = 1/3x \sqrt{x^{2}+5}\), it should be rewritten in a form that enables to apply product and chain rules. Rewrite \(f(x)\) as \(\frac{1}{3}x(x^{2}+5)^{1/2}\). With this format, we can use the product rule (uv)' = u'v + v'u, and within it, the chain rule for \(v = (x^{2}+5)^{1/2}\). Doing this, we find the derivative \(f'(x) = 1/2(x^{2}+5)^{-1/2}2x + (x^{2}+5)^{1/2}\). By simplifying, we get \(f'(x) = x(x^{2}+5)^{-1/2} + (x^{2}+5)^{1/2}\)

Evaluation of the derivative at a given point

Next, evaluate \(f'(x)\) at the given point \(x=2\). This gives the slope of the tangent line at that point. Using the given point \(x=2\) we have \(f'(2) = 2\cdot2(2^{2}+5)^{-1/2} + (2^{2}+5)^{1/2} = 8/3\)

Calculation of the equation of the tangent line

Now use the point-slope form equation \(y - y_{1} = m(x - x_{1})\) where \((x_{1} , y_{1})\) is a point on the line, and \(m\) is the slope of the line. Here, \(m = f'(2) = 8/3\) and \((x_{1} , y_{1}) = (2, 2)\). So, the equation of the tangent line is \(y - 2 = 8/3(x-2)\)

Key concepts

Derivative of a Function

Understanding how to find the derivative of a function is crucial in mathematics, particularly in calculus. The derivative represents the rate at which a function is changing at any given point. It's like a snapshot of a moving object's speed at a particular instant. To find this rate, we use various rules and techniques that let us calculate how steeply (or shallowly) the graph of the function is tilting at a certain point.

For the function in question, the one-third multiplied by the square root of the sum of the square of x and five, we find its steepness at a specific point by taking its derivative. We essentially want to know 'how much does the function value change if we nudge x a little bit?' This concept is the foundation of finding the equation of a tangent line, since it gives us the slope of the line that just 'touches' the graph at one point without crossing through it.

Product Rule

When we have a function that is the product of two or more functions, like \( f(x) = uv \) where both \( u \) and \( v \) are functions of \( x \), we can't just derive each part on its own. We need a special technique called the product rule. The product rule states that the derivative of such a function is \( u'v + uv' \), that is, the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second function.

For instance, with our example function \( f(x) = \frac{1}{3}x(x^{2}+5)^{1/2} \), \( u \) is \( \frac{1}{3}x \) and \( v \) is \( (x^{2}+5)^{1/2} \) which yields the need for the product rule to find the derivative correctly. It helps us to break down more complex equations into simpler parts that we can handle.

Chain Rule

Sometimes, functions are composed of other functions, and this is where the chain rule comes into play. It's a formula for computing the derivative of a composite function. Imagine a function inside another function--this is typical in calculus, and we must tackle it step by step in a chain-like fashion. The chain rule tells us to take the derivative of the outer function and multiply it by the derivative of the inner function.

With our given function, \( (x^{2}+5)^{1/2} \) is an inner function nested inside the square root. We recognize this as our \( v \) from the product rule, and now we have to derive it using the chain rule. The outer function is the square root, and the inner function is \( x^2+5 \). Applying the chain rule, we obtain part of the derivative necessary to find the slope of our tangent line.

Point-Slope Form Equation

After determining the derivative, which gives us the slope at a particular point, we can write the equation for the tangent line. The point-slope form is the go-to formula here. It's expressed as \( y - y_{1} = m(x - x_{1}) \), where \( (x_{1}, y_{1}) \) is the point on the tangent line (the point where the line touches the graph), and \( m \) is the slope at that point.

In our case, the point-slope form becomes \( y - 2 = \frac{8}{3}(x-2) \). This equation means that at the point (2, 2), there's a line that perfectly mirrors how the curve of the function behaves, but only at that one spot; away from that point, the function continues to curve away from the line. The point-slope form is valuable because it allows us to immediately write an equation of a line once we have a point and a slope, which we usually find through differentiation.