Differentiation 1

To Calculate $A$ :- $x^{2} + y^{2} = 4$ Differentiating w.r.t. $x$ we get:- $2x + 2y \dfrac{dy}{dx}=0$ Put $(\sqrt{2},\sqrt{2})$ in above equation:- $\sqrt{2} +\sqrt{2}A=0$ $\implies A = -1$ To Calculate $B$ :- $\sin x + \sin y = \sin x .\sin y$ It can be written as :- $(\sin x-1) (\sin y -1 )= 1$ Differentiating w.r.t. $x$ we get:- $\cos x(\sin y-1) + \cos y(\sin x -1) \dfrac{dy}{dx} = 0$ Put $(\pi,\pi)$ in above equation we get:- $(-1) (0-1) + (-1) (-1) B =0$ $\implies B = -1$ To calculate $C$ :- $2e^{xy} + e^{x} e^{y} - e^{x} - e^{y}= e^{xy+1}$ Above equation can be rewritten as :- $e^{xy}(2-e) + (e^{x} -1)(e^{y} -1) =1$ Differentiating w.r.t. $x$ we get:- $(2-e)e^{xy}(y + x\dfrac{dy}{dx} ) + e^{x}(e^{y}-1) + e^{y} (e^{x} -1)\dfrac{dy}{dx} = 0$ Put $(1,1)$ in the above equation:- $(2-e)e(1+C) + e(e-1) + e(e-1)C =0$ $\implies C=-1$ $\therefore (A+B+C) = -3$

$A$ can be easily visualised as $-1$ by tangent to a circle as in the equation....