Question

find dy/dx of √x+√y=4

Answer 1

dy/dx =-( 4/root x - 1)

Answer 2

Answer:

$\frac{dy}{dx}=1-\frac{4}{\sqrt{x}}$

Step-by-step explanation:

In the question,

We have to find the first derivative of the equation,

√x + √y = 4

So,

For differentiating the function w.r.t 'x' we first, will have to convert the equation in the complete function of 'x'.

On squaring both sides we get,

$\sqrt{x}+\sqrt{y}=4\\\sqrt{y}=4-\sqrt{x}\\On\ squaring\ both\ sides\ we,\ get,\\y=(4-\sqrt{x})^{2}\\y=16+x-8\sqrt{x}$

Now, on differentiating the function w.r.t 'x' we get,

$y=16+x-8\sqrt{x}\\\frac{dy}{dx}=1-\frac{8}{2\sqrt{x}}\\\frac{dy}{dx}=1-\frac{4}{\sqrt{x}}$

Therefore, the first derivative of the equation is given by,

$\frac{dy}{dx}=1-\frac{4}{\sqrt{x}}$