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.