Find y' for sqrt(x)+ sqrt(y)=4, and evaluate y'(4)
Find $y'$ for $\sqrt{x}+\sqrt{y}=4$, and evaluate $y'(4)$.
Explanation of Method
A parabolic arc joining two axis points, with slope -1 at its midpoint.
Step 1: Differentiate both sides with respect to $x$
\[ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}\,y' = 0. \]
Step 2: Solve for $y'$
The factors $\tfrac12$ cancel, and the reciprocals invert: $y' = -\frac{\sqrt{y}}{\sqrt{x}} = -\sqrt{\frac{y}{x}}$.
Step 3: Find $y$ when $x = 4$
\[ \sqrt4+\sqrt y = 4 \;\Longrightarrow\; 2+\sqrt y = 4 \;\Longrightarrow\; \sqrt y = 2 \;\Longrightarrow\; y = 4. \]
Step 4: Evaluate
\[ y'(4) = -\sqrt{\frac44} = -1. \]
Step 5: Verification
Solving explicitly, $y = \bigl(4-\sqrt x\bigr)^{2}$, and differentiating directly: \[y' = 2\bigl(4-\sqrt x\bigr)\cdot\left(-\frac{1}{2\sqrt x}\right) = -\frac{4-\sqrt x}{\sqrt x} = -\frac{\sqrt y}{\sqrt x} \;\;\checkmark\]
\[x = 1:\quad y = 9,\ y' = -3; \qquad x = 9:\quad y = 1,\ y' = -\tfrac13 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = -\sqrt{\dfrac{y}{x}}, \qquad y'(4) = -1\,}\]
Remark
Remark. The curve runs from $(16,0)$ to $(0,16)$ and is symmetric in $x$ and $y$; the point $(4,4)$ is where it crosses the line $y = x$, so the slope $-1$ there is exactly what the symmetry demands. Squaring the relation shows it is an arc of a parabola with axis along $y = x$, not a straight line — the slope varies from $0$ to $-\infty$ across it.