Finding the Derivative of a Function

We are asked to find the derivative of the function:

\(y = \frac{x \sqrt{x^2−16}}{2} − 8 \ln\left|x + \sqrt{x^2−16}\right|\)

To find \(\frac{\text{dy}}{\text{dx}}\), we will differentiate each term with respect to \(x\). Let the first term be \(u = \frac{x \sqrt{x^2−16}}{2}\) and the second term be \(v = 8 \ln\left|x + \sqrt{x^2−16}\right|\). Then \(y = u - v\), and \(\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}\).

Differentiating the First Term: \(\frac{\text{d}}{\text{dx}}\left(\frac{\text{x} \sqrt{\text{x}^2−16}}{2}\right) \)

We can write \(u = \frac{1}{2} x (x^2-16)^{1/2}\). We will use the product rule \((fg)' = f'g + fg'\) and the chain rule.

Let \(f = x\) and \(g = (x^2-16)^{1/2}\).

• Derivative of \(f\): \(f' = \frac{d}{dx}(x) = 1\)
• Derivative of \(g\): \(g' = \frac{d}{dx}((x^2-16)^{1/2})\). Using the chain rule, this is \(\frac{1}{2}(x^2-16)^{-1/2} \cdot \frac{d}{dx}(x^2-16)\).
  \(\frac{d}{dx}(x^2-16) = 2x\).
  So, \(g' = \frac{1}{2}(x^2-16)^{-1/2} \cdot 2x = x(x^2-16)^{-1/2} = \frac{x}{\sqrt{x^2-16}}\).

Now apply the product rule for \(x \sqrt{x^2-16}\):

\(\frac{d}{dx}(x \sqrt{x^2-16}) = f'g + fg' = 1 \cdot \sqrt{x^2-16} + x \cdot \frac{x}{\sqrt{x^2-16}}\)

\(= \sqrt{x^2-16} + \frac{x^2}{\sqrt{x^2-16}}\)

To combine these terms, find a common denominator:

\(= \frac{(\sqrt{x^2-16})^2}{\sqrt{x^2-16}} + \frac{x^2}{\sqrt{x^2-16}} = \frac{x^2-16 + x^2}{\sqrt{x^2-16}} = \frac{2x^2-16}{\sqrt{x^2-16}}\)

Now, we multiply by the constant \(\frac{1}{2}\) for \(\frac{du}{dx}\):

\(\frac{du}{dx} = \frac{1}{2} \cdot \frac{2x^2-16}{\sqrt{x^2-16}} = \frac{x^2-8}{\sqrt{x^2-16}}\)

Differentiating the Second Term: \(\frac{\text{d}}{\text{dx}}\left(8 \ln\left|x + \sqrt{x^2−16}\right|\right)\)

We will use the chain rule and the fact that \(\frac{d}{dw}(\ln|w|) = \frac{1}{w}\).

Let \(w = x + \sqrt{x^2-16}\). The derivative is \(8 \cdot \frac{d}{dw}(\ln|w|) \cdot \frac{dw}{dx}\).

• \(\frac{d}{dw}(\ln|w|) = \frac{1}{w} = \frac{1}{x + \sqrt{x^2-16}}\)
• \(\frac{dw}{dx} = \frac{d}{dx}(x + \sqrt{x^2-16}) = \frac{d}{dx}(x) + \frac{d}{dx}(\sqrt{x^2-16})\)
  \(\frac{d}{dx}(x) = 1\).
  \(\frac{d}{dx}(\sqrt{x^2-16})\) was calculated before as \(\frac{x}{\sqrt{x^2-16}}\).
  So, \(\frac{dw}{dx} = 1 + \frac{x}{\sqrt{x^2-16}}\)
• Combine terms for \(\frac{dw}{dx}\): \(1 + \frac{x}{\sqrt{x^2-16}} = \frac{\sqrt{x^2-16}}{\sqrt{x^2-16}} + \frac{x}{\sqrt{x^2-16}} = \frac{\sqrt{x^2-16} + x}{\sqrt{x^2-16}}\)

Now, combine these parts for \(\frac{dv}{dx}\):

\(\frac{dv}{dx} = 8 \cdot \frac{1}{x + \sqrt{x^2-16}} \cdot \frac{x + \sqrt{x^2-16}}{\sqrt{x^2-16}}\)

Notice that the term \((x + \sqrt{x^2-16})\) cancels out:

\(\frac{dv}{dx} = 8 \cdot \frac{1}{\sqrt{x^2-16}} = \frac{8}{\sqrt{x^2-16}}\)

Combining the Derivatives to Find \(\frac{\text{dy}}{\text{dx}}\)

Now we calculate \(\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}\):

\(\frac{dy}{dx} = \frac{x^2-8}{\sqrt{x^2-16}} - \frac{8}{\sqrt{x^2-16}}\)

Since the denominators are the same, we can combine the numerators:

\(\frac{dy}{dx} = \frac{(x^2-8) - 8}{\sqrt{x^2-16}} = \frac{x^2-16}{\sqrt{x^2-16}}\)

For \(x^2 > 16\), we know that \(x^2-16 = (\sqrt{x^2-16})^2\). So we can simplify:

\(\frac{dy}{dx} = \frac{(\sqrt{x^2-16})^2}{\sqrt{x^2-16}} = \sqrt{x^2-16}\)

Thus, the derivative of the given function is \(\sqrt{x^2-16}\).

Revision Table: Differentiation Techniques

Additional Information on Derivatives and Hyperbolic Functions

The structure of the function \( \ln\left|x + \sqrt{x^2−a^2}\right| \) is related to the inverse hyperbolic cosine function. Specifically, for \(x > a\), \(\text{arccosh}(x/a) = \ln\left|x + \sqrt{x^2−a^2}\right| - \ln|a|\). The derivative of \(\text{arccosh}(x/a)\) with respect to \(x\) is \(\frac{1}{\sqrt{(x/a)^2 - 1}} \cdot \frac{1}{a} = \frac{1}{\frac{\sqrt{x^2-a^2}}{a}} \cdot \frac{1}{a} = \frac{a}{\sqrt{x^2-a^2}} \cdot \frac{1}{a} = \frac{1}{\sqrt{x^2-a^2}}\). In our problem, \(a=4\), so the derivative of \(\ln\left|x + \sqrt{x^2−16}\right|\) is indeed \(\frac{1}{\sqrt{x^2-16}}\). This confirms the result obtained using direct differentiation rules.

The first term \(\frac{x \sqrt{x^2−16}}{2}\) is similar in structure to terms that arise when integrating functions involving \(\sqrt{x^2-a^2}\) using trigonometric substitution or hyperbolic substitution. The derivative calculation involves standard rules like product rule and chain rule, which are fundamental in calculus for finding instantaneous rates of change.