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The given equation is:

\[
y = \tan^{-1}\left(\frac{5}{x}\right)
\]

We are asked to find \(\frac{dy}{dx}\).

To solve for \(\frac{dy}{dx}\), we need to differentiate the equation with respect to \(x\). The derivative of \(\tan^{-1}(u)\) is:

\[
\frac{d}{dx} \left(\tan^{-1}(u)\right) = \frac{1}{1+u^2} \cdot \frac{du}{dx}
\]

Here, \(u = \frac{5}{x}\). Now let's differentiate:

1. First, apply the chain rule:
\[
\frac{dy}{dx} = \frac{1}{1 + \left(\frac{5}{x}\right)^2} \cdot \frac{d}{dx}\left(\frac{5}{x}\right)
\]

2. Now, differentiate \(\frac{5}{x}\):
\[
\frac{d}{dx} \left(\frac{5}{x}\right) = -\frac{5}{x^2}
\]

3. Substitute this result back into the equation:
\[
\frac{dy}{dx} = \frac{1}{1 + \frac{25}{x^2}} \cdot \left(-\frac{5}{x^2}\right)
\]

4. Simplify the expression:
\[
\frac{dy}{dx} = \frac{-\frac{5}{x^2}}{1 + \frac{25}{x^2}}
\]

5. Finally, multiply both the numerator and denominator by \(x^2\) to get rid of the fractions:
\[
\frac{dy}{dx} = \frac{-5}{x^2 + 25}
\]

Thus, the derivative is:

\[
\boxed{\frac{-5}{x^2 + 25}}
\]

Would you like more details or have any other questions?

Here are 5 related questions to expand on this concept:
1. How does the chain rule apply to inverse trigonometric functions in general?
2. What is the derivative of \(\tan^{-1}(x)\) itself, without any substitution?
3. How would the solution change if the function were \(y = \tan^{-1}\left(\frac{a}{x}\right)\) for a general constant \(a\)?
4. Can you find the second derivative, \(\frac{d^2y}{dx^2}\), for this function?
5. How would you apply this differentiation technique to other inverse functions like \(\sin^{-1}(x)\)?

**Tip**: When differentiating functions involving inverse trigonometric functions, always use the chain rule carefully, especially when the argument is a more complex expression.

Learn how to find the derivative of y = tan^(-1)(5/x) with a detailed step-by-step solution. This problem involves inverse trigonometric functions and the chain rule, suitable for students in Grades 11-12.

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