Question

if $sqrt{x}+sqrt{y}=8$ and $y(9) = 25$, find $y(9)$ by implicit differentiation.

Explanation:

Step1: Differentiate both sides of the equation

Differentiate $\sqrt{x}+\sqrt{y}=8$ with respect to $x$. The derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$, and using the chain - rule, the derivative of $\sqrt{y}$ with respect to $x$ is $\frac{1}{2\sqrt{y}}y'$. The derivative of the constant 8 is 0. So we have $\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}y' = 0$.

Step2: Solve for $y'$

First, subtract $\frac{1}{2\sqrt{x}}$ from both sides: $\frac{1}{2\sqrt{y}}y'=-\frac{1}{2\sqrt{x}}$. Then multiply both sides by $2\sqrt{y}$ to get $y'=-\frac{\sqrt{y}}{\sqrt{x}}$.

Step3: Substitute $x = 9$ and $y(9)=25$

When $x = 9$ and $y = 25$, we substitute these values into the formula for $y'$. So $y'(9)=-\frac{\sqrt{25}}{\sqrt{9}}=-\frac{5}{3}$.

Answer:

$-\frac{5}{3}$