homework7: problem 3\n(2 points)\nfor the equation given below, evaluate $\\frac{dy}{dx}$ at the point $(25…

homework7: problem 3\n(2 points)\nfor the equation given below, evaluate $\\frac{dy}{dx}$ at the point $(25, 16)$.\n$\\sqrt{x}+\\sqrt{y}=9$\n$\\frac{dy}{dx}$ at $(25, 16)=$\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.\nemail instructor\npage generated october 19, 2025, 8:37:42 pm cdt\nwebwork © 1996 - 2024 | theme: math4 - ttu | ww_version 2.19 | pg_version 2.19\nthe webwork project

homework7: problem 3\n(2 points)\nfor the equation given below, evaluate $\\frac{dy}{dx}$ at the point $(25, 16)$.\n$\\sqrt{x}+\\sqrt{y}=9$\n$\\frac{dy}{dx}$ at $(25, 16)=$\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.\nemail instructor\npage generated october 19, 2025, 8:37:42 pm cdt\nwebwork © 1996 - 2024 | theme: math4 - ttu | ww_version 2.19 | pg_version 2.19\nthe webwork project

Answer

Explanation:

Step1: Differentiate both sides with respect to (x)

Differentiate (\sqrt{x}+\sqrt{y}=9). Using the power rule ((x^n)^\prime = nx^{n - 1}) and the chain - rule ((y^n)^\prime=ny^{n - 1}\frac{dy}{dx}), we have (\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0).

Step2: Solve for (\frac{dy}{dx})

Subtract (\frac{1}{2\sqrt{x}}) from both sides: (\frac{1}{2\sqrt{y}}\frac{dy}{dx}=-\frac{1}{2\sqrt{x}}). Then multiply both sides by (2\sqrt{y}) to get (\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}).

Step3: Substitute (x = 25) and (y = 16)

Substitute (x = 25) (so (\sqrt{x}=5)) and (y = 16) (so (\sqrt{y}=4)) into (\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}). We get (\frac{dy}{dx}=-\frac{4}{5}).

Answer:

(-\frac{4}{5})