Math, asked by vishwanathview2288, 1 year ago

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

Answers

Answered by Sharad001
88

★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} }

Answered by amitnrw
0

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)"

brainly.in/question/15286193

sin (ax+b) फलन का अवकलन कीजिए

brainly.in/question/15286166

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