Question

Find dy/dx

if x² + y2 + xy-y=0 at(1,2)​

Answer 1

Step-by-step explanation:

x² + y² + xy - y = 0

D.w.r.to.x

2x + 2ydy/dx + xdy/dx + y - dy/dx = 0

2x + y + (2y+x-1)dy/dx = 0

dy/dx = -(2x+y)/(2y+x-1)

at point (1,2)

dy/dx = -(2+2)/(4+1-1) = -4/4 = -1

Answer 2

The dy/dx will be -1

Given : x²+y²+xy-y = 0 at (1,2)

To find : The value of dy/dx

Solution :

We can simply solve this mathematical problem by using the following mathematical process. (our goal is to find dy/dx)

Now,

x²+y²+xy-y = 0

Differentiating both sides w.r.t. x :

$2x + 2y \frac{dy}{dx} + [ \frac{d}{dx}(x) \times y \: + x \times \frac{d}{dx} (y)] - \frac{dy}{dx} = 0$

$2x + 2y \frac{dy}{dx} + y + x \frac{dy}{dx} - \frac{dy}{dx} = 0$

$2y \frac{dy}{dx} + x \frac{dy}{dx} - \frac{dy}{dx} = - 2x - y$

$\frac{dy}{dx}( 2y + x - 1 ) = - 2x - y$

$\frac{dy}{dx} = \frac{ - 2x - y}{2y + x - 1 }$

Now,

$\frac{dy}{dx}|_{(1,2)} = \frac{ -[ (2 \times 1) + 2]}{(2 \times 2) + 1 - 1 }$

$\frac{dy}{dx}|_{(1,2)} = \frac{ - 4}{4}$

$\frac{dy}{dx}|_{(1,2)} = - 1$

(This will be considered as the final result.)

Hence, dy/dx will be -1