Recall that if \( n \) is any real number and \( u=g(x) \) is differentiable, then we have the following. \[ \frac{d}{d x}\left(u^{n}\right)=n u^{n-1} \frac{d u}{d x} \] If we let \( u=\frac{x}{x+9} \) and \( n=\frac{1}{2} \), then we can write our given function in the form \( y=u^{n} \), allowing us to use the derivative rule above. Before applying it, we must first find \( \frac{d u}{d x} \). Doing so requires the quotient rule and gives the following result. \[ u=\frac{x}{x+9} \] \( \frac{d u}{d x}=\frac{(x+1)(1)-x(1)}{(x+9)^{2}} \)

To compute the derivative of \( u = \frac{x}{x+9} \), we start with the quotient rule, which states that if you have a function \( u = \frac{f(x)}{g(x)} \), then its derivative is given by \( \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \). In this case, \( f(x) = x \) and \( g(x) = x + 9 \). Thus, \( f'(x) = 1 \) and \( g'(x) = 1 \). Plugging into the quotient rule formula, we find: \[ \frac{d u}{d x}=\frac{(x + 9)(1) - (x)(1)}{(x + 9)^2} = \frac{x + 9 - x}{(x + 9)^2} = \frac{9}{(x + 9)^2}. \] So, the derivative of \( u \) becomes \( \frac{d u}{d x} = \frac{9}{(x+9)^2} \). Now, with \( u = \frac{x}{x+9} \) and \( n = \frac{1}{2} \), we can apply the power rule as follows: \[ \frac{d}{dx}\left(u^{\frac{1}{2}}\right) = \frac{1}{2} u^{-\frac{1}{2}} \frac{d u}{d x}. \] This means we can now substitute the value of \( u \) and \( \frac{d u}{d x} \) to compute the derivative of \( y \)! Now, here's a fun fact: The quotient rule is like a culinary recipe; you need to be precise with each ingredient. Just as a pinch of salt can enhance a dish, the correct use of derivative rules can elevate your calculus skills to the next level! And speaking about practice, if you ever feel that derivatives are overwhelming, try eyeing a graph of the function you're working on. Visualizing how the function behaves provides great intuition and can help clarify your calculations—definitely a foundational move for anyone delving into calculus!