Final answer:
By differentiating the given equation implicitly, substituting x with 1 and y with 36, we find that the derivative of y at x equals 1 is y'(1) = -6.
Step-by-step explanation:
We are given that \(\sqrt{x} + \sqrt{y} = 7\) and \(y(1) = 36\). We need to find \(y'\) at \(x = 1\) using implicit differentiation. First, let's differentiate both sides of the equation with respect to x:
\(\frac{d}{dx}(\sqrt{x} + \sqrt{y}) = \frac{d}{dx}(7)\)
\(\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\cdot y' = 0\)
Since \(y(1) = 36\), we can substitute 1 for x and 36 for y into the differentiated equation and solve for \(y'\):
\(\frac{1}{2\sqrt{1}} + \frac{1}{2\sqrt{36}}\cdot y'(1) = 0\)
\(\frac{1}{2} + \frac{1}{12}y'(1) = 0\)
\(y'(1) = -6\)
Thus, \(y'(1) = -6\), which is the rate of change of y with respect to x at the point where x equals 1.