Question

√x+√y=1 differanciate with respect to x
$\sqrt{x } + \sqrt{y} = 1$
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Answer 1

Question :

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

differnatiate with respect to x

Theory :

chain rule :

let y = f(t) and t = g(x)

then ,

$\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}$

Formulas

$\frac{d(x {}^{n} )}{dx} = nx {}^{n - 1}$

$\frac{d(constant)}{dx} = 0$

Solution :

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

differnatiate with respect to x ,

$\frac{d(x {}^{ \frac{1}{2} }) }{dx} + \frac{d(y {}^{ \frac{1}{2} }) }{dy} \times \frac{dy}{dx} = 0$

$\implies \: \frac{1}{2} x {}^{ \frac{1}{2} - 1 } + \frac{1}{2} y {}^{ \frac{1}{2} - 1} \times \frac{dy}{dx} = 0$

$\implies \: \frac{1}{2 \sqrt{x} } + \frac{1}{2 \sqrt{y} } \times \frac{dy}{dx} = 0$

$\implies \: \frac{1}{2 \sqrt{y} } \times \frac{dy}{dx} = - \frac{1}{2 \sqrt{x} }$

$\implies \: \frac{dy}{dx} = \frac{ - 2 \sqrt{y} }{2 \sqrt{x} }$

$\frac{dy}{dx} = - \frac{ \sqrt{y} }{ \sqrt{x} }$

which is the required solution!