Question

if 13x²+2x²y+y³=1.find dy/dx at (1,-2)​

Answer 1

${\bold{\huge{\red{\underline{\green{ANSWER}}}}}}$

$13 {x}^{2} + 2 {x}^{2} y + {y}^{3}$

differentiating both the side w.r.t. X

we get,

$26x + 4xy + 2 {x}^{2} \frac{dy}{dx} + 2 {y}^{2} \frac{dy}{dx} = 0$

$26x + 4xy = (- 2 {x}^{2} - 2 {y}^{2} ) \frac{dy}{dx}$

$\frac{26x + 4xy}{ - 2 {x}^{2} - 2 {y}^{2} } = \frac{dy}{dx}$

at (1,-2)

$\frac{26 - 8}{ - 2 - 2} = \frac{dy}{dx}$

$\frac{18}{ - 4} = \frac{dy}{dx}$

Answer 2

Answer:

if 13x²+2x²y+y³=1.find dy/dx at (1,-2)