Question

if $\sqrt{x}+\sqrt{y}=11$ and $y(25)=36$, find $y(25)$ by implicit differentiation.

Explanation:

Step1: Differentiate both sides of the equation

Differentiate \(\sqrt{x}+\sqrt{y}=11\) with respect to \(x\).
Using the power rule \((x^n)^\prime = nx^{n - 1}\), we have \(\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}y^\prime=0\).

Step2: Solve for \(y^\prime\)

Multiply through by \(2\) to get \(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}y^\prime = 0\).
Then isolate \(y^\prime\): \(\frac{1}{\sqrt{y}}y^\prime=-\frac{1}{\sqrt{x}}\), so \(y^\prime=-\frac{\sqrt{y}}{\sqrt{x}}\).

Step3: Substitute \(x = 25\) and \(y=36\)

When \(x = 25\) and \(y = 36\), \(y^\prime=-\frac{\sqrt{36}}{\sqrt{25}}\).

Answer:

\(-\frac{6}{5}\)