Question

if √x+√y=1, find dy/dx at (4,9)​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Given :

$\sqrt{x}+\sqrt{y}=1$ at (4,9)

To find : $\dfrac{dy}{dx}=?$

Solution :

$\sqrt{y}=1-\sqrt{x}$

Differentiating w.r.t x, we get :

$\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}=-\dfrac{1}{2\sqrt{x}}\\\\At\ (4,9),\\\\\dfrac{1}{2\sqrt{9}}\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{4}}\\\\\dfrac{1}{2\times 3}\dfrac{dy}{dx}=\dfrac{1}{2\times 2}\\\\\dfrac{1}{6}\dfrac{dy}{dx}=\dfrac{1}{4}\\\\\dfrac{dy}{dx}=\dfrac{6}{4}=\dfrac{3}{2}$

Hence, the value of $\dfrac{dy}{dx}=\dfrac{3}{2}$