Question

homework7: problem 3
(2 points)
for the equation given below, evaluate $\frac{dy}{dx}$ at the point $(25, 16)$.
$\sqrt{x}+\sqrt{y}=9$
$\frac{dy}{dx}$ at $(25, 16)=$
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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:

\(-\frac{4}{5}\)