What is the process to solve \sqrt{x}+\sqrt{y}=5 for the point (4,9)?

[riri]

Hello!
Can someone help me with the process of solving
$$\sqrt{x}$$+$$\sqrt{y}$$=5 on point (4,9)?
With implicit, I differntiated both sides and ended up with 1/2x^-1/2+1/2y^-1/2$$\d{y}{x}$$=0
and I tried to isolate the dy/dx, but how do I get rid of the others?

And with explicit, I isolated y to one side and got y=5-$$\sqrt{x}$$^2
and don't know how how to do this

THANK YOU!

 

[MarkFL]

We are given the implicit relation:

$$\sqrt{x}+\sqrt{y}=5\tag{1}$$

and are asked to find:

$$\left.\d{y}{x}\right|_{(x,y)=(4,9)}$$

So, implicitly differentiating (1) w.r.t $x$, we obtain:

$$\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\cdot\d{y}{x}=0$$

Multiply through by $2$:

$$\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\cdot\d{y}{x}=0$$

Subtract through by $$\frac{1}{\sqrt{x}}$$

$$\frac{1}{\sqrt{y}}\cdot\d{y}{x}=-\frac{1}{\sqrt{x}}$$

Multiply through by $$\sqrt{y}$$:

$$\d{y}{x}=-\frac{\sqrt{y}}{\sqrt{x}}$$

Hence:

$$\left.\d{y}{x}\right|_{(x,y)=(4,9)}=-\frac{\sqrt{9}}{\sqrt{4}}=-\frac{3}{2}$$

Now, if we wish to check our answer by doing it explicitly, we would write (1) as:

$$y=25-10\sqrt{x}+x\tag{2}$$

Thus:

$$\d{y}{x}=0-\frac{10}{2\sqrt{x}}+1=1-\frac{5}{\sqrt{x}}$$

And so we find:

$$\left.\d{y}{x}\right|_{x=4}=1-\frac{5}{\sqrt{4}}=1-\frac{5}{2}=-\frac{3}{2}$$