Question

x^{3} + x^{2}y + xy^{2} + y^{3} = 81 dy/dx ज्ञात कीजिए

Answer 1

★彡 Q u e s t i o n ★彡

$\rm if \: {x}^{3} + {x}^{2} y + {y}^{3} = 81 \: then \: find \: \frac{dy}{dx}$

★彡 A n s w e r 彡★

$\to \boxed{ \rm \frac{dy}{dx} = \frac{ - (3 {x}^{2} + 2xy)}{ {x}^{2} + 3y^{2} } } \:$

★彡 S o l u t i o n 彡★

We have ,

$\mapsto \rm {x}^{3} + {x}^{2} y + {y}^{3} = 81 \\ \\ \rm \red{ differentiate \: with \: respect \: to} \: x \: on \: \\ \green{ \rm both \: sides} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \rm \frac{d}{dx} {x}^{n} = n {x}^{n - 1} } \\ \\ \large \boxed{ \bf \: chain \: rule \: } \\ \\ \to \rm \frac{d}{dx} uv = u \frac{d}{dx} v + v \frac{d}{dx} u \\ \\ \sf \: therefore \\ \\ \to \rm 3 {x}^{2} + {x}^{2} \frac{dy}{dx} + y \frac{d}{dx} {x}^{2} + 3y^{2}\frac{dy}{dx} = 0 \\ \\ \to \rm \rm 3 {x}^{2} + {x}^{2} \frac{dy}{dx} + 2xy + 3y^{2} \frac{dy}{dx} = 0 \\ \\ \to \rm \frac{dy}{dx} \bigg( {x}^{2} + 3y^{2} \bigg) = - 3 {x}^{2} - 2xy \\ \\ \to \boxed{ \rm \frac{dy}{dx} = \frac{ - (3 {x}^{2} + 2xy)}{ {x}^{2} + 3y^{2}} }$

Answer 2

(dy/dx) = - (3x²  + 2xy + y²)/(x² + 2xy + 3y²)   यदि x³ + x²y  + xy² + y³   = 81

Step-by-step explanation:

dy/dx ज्ञात कीजिए

x³ + x²y  + xy² + y³   = 81

=> 3x²  + x²dy/dx  + y2x  + x2ydy/dx  + y² + 3y²dy/dx  = 0

=> (dy/dx)(x² + 3y²  + 2xy)  + 3x²  + 2xy + y² = 0

=>  (dy/dx) = - (3x²  + 2xy + y²)/(x² + 2xy + 3y²)

और अधिक जानें :

sin(x²+5)"

brainly.in/question/15286193

sin (ax+b) फलन का अवकलन कीजिए

brainly.in/question/15286166