Question

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

Answer 1

★彡 QuesTion 彡★

$\sf if \: xy + {y}^{2} = \tan x + y \: then \: find \: \frac{dy}{dx}$

★彡 AnsWer 彡★

$\to \boxed{ \sf \frac{dy}{dx} = \frac{ { \sec}^{2}x + y }{x + 2y - 1} } \:$

★彡 SoluTion 彡★

We have ,

$\leadsto \sf xy + {y}^{2} = \tan x + y \\ \\ \rm \red{ differentiate \: with \: }respect \: to \: x \\ \large\underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \sf here \: we \: use \: chain \: rule \: of \: differentiation \\ \rm that \: is \: - \\ \to \sf \frac{d}{dx} uv = u \frac{d}{dx} v + v \frac{d}{dx} u \\ \large\underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \: \\ \therefore \\ \to \sf x \frac{dy}{dx} + y \frac{d}{dx} x + 2y \frac{dy}{dx} = \frac{d}{dx} \tan x + \frac{dy}{dx} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \sf \frac{d}{dx} {x}^{n} = n {x}^{n - 1} \: and \: \frac{d}{dx} = { \sec}^{2} x} \\ \\ \to \sf x \frac{dy}{dx} + y + 2y \frac{dy}{dx} = { \sec}^{2} x + \frac{dy}{dx} \\ \\ \to \sf x \frac{dy}{dx} + 2y \frac{dy}{dx} - \frac{dy}{dx} = { \sec}^{2} x - y \\ \\ \to \sf \frac{dy}{dx} \{x + 2y - 1 \} = { \sec}^{2} x + y \\ \\ \to \boxed{ \sf \frac{dy}{dx} = \frac{ { \sec}^{2}x + y }{x + 2y - 1} }$

Answer 2

dy/dx = (Sec²x - y)/(x + 2y  - 1)  xy + y² = tan x + y

Step-by-step explanation:

dy/dx ज्ञात कीजिए

xy + y²  = tanx + y

=> x dy/dx  + y  + 2ydy/dx  = Sec²x   +  dy/dx

=> (dy/dx)(x + 2y  - 1)  = Sec²x - y

=> dy/dx = (Sec²x - y)/(x + 2y  - 1)

और अधिक जानें :

sin(x²+5)"

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