Question

find dy/dx if:

sqrt(1-x^2) + sqrt(1-y^2) = a(x-y)

Answer 1

Question :-

$\bf{Find \: \frac{dy}{dx} \: if - } \\ \to \bf{ \sqrt{1 - {x}^{2} } + \sqrt{1 - {y}^{2} } =a (x - y)}$

Answer :-

$\to \boxed{\bf{ \frac{dy}{dx} = \frac{ \bigg(a + \frac{x}{ \sqrt{1 - {x}^{2} } } \bigg)}{ \bigg(a - \frac{y}{ \sqrt{1 - {y}^{2} } } \bigg) } }} \\ \sf{or} \\ \: \to \boxed{ \bf{ \frac{dy}{dx} = \frac{ \sqrt{1 - {y}^{2} } \:\bigg( a \sqrt{1 - {x}^{2} } + x \bigg) \: }{ \sqrt{1 - {x}^{2} } \bigg(a \sqrt{1 - {y}^{2} } - y \bigg) } }} \:$

To Find :-

$\implies \: \bf{ \frac{dy}{dx} }$

Explanation :-

We have

$\to \bf{ \sqrt{1 - {x}^{2} } + \sqrt{1 - {y}^{2} } =a (x - y)} \:$

differentiate with respect to x on both sides .

$\because \boxed{\bf{\frac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } }} \\ \therefore \\ \to \bf{ \frac{1}{2 \sqrt{1 - {x}^{2} } } \frac{d}{dx} (1 - {x}^{2} ) + \frac{1}{2 \sqrt{1 - {y}^{2} } } \frac{d}{dx} (1 - {y)}^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf{ = a - a \frac{dy}{dx} } \\ \\ \to \bf{} \frac{ - 2x}{2 \sqrt{1 - {x}^{2} } } - \frac{2y}{2 \sqrt{1 - {y}^{2} } } \frac{dy}{dx} = a - a \frac{dy}{dx} \\ \\ \to \bf{ a \frac{dy}{dx} - \frac{y}{ \sqrt{1 - {y}^{2} } } \frac{dy}{dx} = a + \frac{x}{ \sqrt{1 - {x}^{2} } }} \\ \\ \to \bf{ \frac{dy}{dx} \bigg(a - \frac{y}{ \sqrt{1 - {y}^{2} } } \bigg) = a + \frac{x}{ \sqrt{1 - {x}^{2} } } } \\ \\ \to \boxed{\bf{ \frac{dy}{dx} = \frac{ \bigg(a + \frac{x}{ \sqrt{1 - {x}^{2} } } \bigg)}{ \bigg(a - \frac{y}{ \sqrt{1 - {y}^{2} } } \bigg) } }} \\ \sf{or}$

$\to \bf{\frac{dy}{dx} = \frac{ \frac{a \sqrt{1 - {x}^{2} } + x}{ \sqrt{1 - {x}^{2} } } }{ \frac{a \sqrt{1 - {y}^{2} - y} }{ \sqrt{1 - {y}^{2} } } } } \\ \\ \to \boxed{ \bf{ \frac{dy}{dx} = \frac{ \sqrt{1 - {y}^{2} } \:\bigg( a \sqrt{1 - {x}^{2} } + x \bigg) \: }{ \sqrt{1 - {x}^{2} } \bigg(a \sqrt{1 - {y}^{2} } - y \bigg) } }}$

Answer 2

$\tt{\huge{\green{Solution_{Sd} \:: - }}}$

$\tt{\orange {Step-By-Step-Explaination \::- }}$>

$\begin{lgathered}\because \boxed{\bf{\dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2 \sqrt{x} } }} \\ \therefore \\ \to \bf{ \dfrac{1}{2 \sqrt{1 - {x}^{2} } } \dfrac{d}{dx} (1 - {x}^{2} ) + \dfrac{1}{2 \sqrt{1 - {y}^{2} } } \dfrac{d}{dx} (1 - {y)}^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf{ = a - a \dfrac{dy}{dx} } \\ \\ \to \bf{} \dfrac{ - 2x}{2 \sqrt{1 - {x}^{2} } } - \dfrac{2y}{2 \sqrt{1 - {y}^{2} } } \dfrac{dy}{dx} = a - a \dfrac{dy}{dx} \\ \\ \to \bf{ a \dfrac{dy}{dx} - \frac{y}{ \sqrt{1 - {y}^{2} } } \dfrac{dy}{dx} = a + \dfrac{x}{ \sqrt{1 - {x}^{2} } }} \\ \\ \to \bf{ \dfrac{dy}{dx} \bigg(a - \dfrac{y}{ \sqrt{1 - {y}^{2} } } \bigg) = a + \dfrac{x}{ \sqrt{1 - {x}^{2} } } } \\ \\ \to \boxed{\bf{ \dfrac{dy}{dx} = \dfrac{ \bigg(a + \dfrac{x}{ \sqrt{1 - {x}^{2} } } \bigg)}{ \bigg(a - \dfrac{y}{ \sqrt{1 - {y}^{2} } } \bigg) } }} \\ \sf{or} \\\end{lgathered}$

$</p><p>\begin{lgathered}\to \bf{d\frac{dy}{dx} = \dfrac{ \dfrac{a \sqrt{1 - {x}^{2} } + x}{ \sqrt{1 - {x}^{2} } } }{ \dfrac{a \sqrt{1 - {y}^{2} - y} }{ \sqrt{1 - {y}^{2} } } } } \\ \\ \to \boxed{ \bf{ \dfrac{dy}{dx} = \dfrac{ \sqrt{1 - {y}^{2} } \:\bigg( a \sqrt{1 - {x}^{2} } + x \bigg) \: }{ \sqrt{1 - {x}^{2} } \bigg(a \sqrt{1 - {y}^{2} } - y \bigg) } }}\end{lgathered}$