Question

Hola ❤

Good morning guys ❤

plzz solve it

❌No spam ❌

​

Answer 1

REFER TO THE ATTACHMENT FOR THE ANSWER

$<marquee$please mark it as brainliest

Answer 2

${ \bold{ \boxed{ \boxed{ \huge\red{ \mathfrak{ \bigstar \: ANSWER}}}}}}$

${ \bold{ \underline{Given} : - }} \\ \\ { \bold{ \orange{xy + {y}^{2} = \tan(x + y) }}}$

${ \bold{ \underline{To \: \: find} : - }}\\ \\ { \bold{ \green{ \huge{ \frac{dy}{dx} }}}} \\ \\ \\ { \bold{ \boxed{ \boxed{ \red{ \huge{ \mathfrak{ \bigstar \: solution \: \bigstar}}}}}}}$

${ \bold{ \orange{ : \implies \:xy + {y}^{2} = \tan(x + y) }}} \\ \\ { \bold{ \blue{ \:now \: \: d.w.r \: \: \: to \: \: \: x \: - }}} \\ \\ \\ { \bold{ \orange{ : \implies \: x \frac{dy}{dx} + y + 2y \frac{dy}{dx} = {sec}^{2} (x + y)(1 + \frac{dy}{dx}) }}} \\ \\ \\ { \bold{ \orange{ : \implies \: \frac{dy}{dx} (x + 2y) + y = {sec}^{2}(x + y) + {sec}^{2} (x + y)( \frac{dy}{dx} ) }}} \\ \\ \\ { \bold{ \orange{ : \implies \: \frac{dy}{dx} (x + 2y - {sec}^{2} (x + y)) = {sec}^{2} (x + y) - y}}} \\ \\ { \bold{ \orange{ : \implies \: \frac{dy}{dx} = \frac{ {sec}^{2}(x + y) - y }{x +2 y - {sec}^{2} (x + y)} }}}$

${ \bold{ \underline{ \red{used \: \: formula} } : - }} \\ \\ { \bold{ \green{ \: \: \: \: \: \: . \: \: \frac{d(ab)}{dx} = \: b \frac{da}{dx} + a \frac{db}{dx} }}} \\ \\ \: \: \: \: \: \: { \bold{ \green{ . \: \: \: \frac{d( tanx) }{dx} = {sec}^{2}(x) }}}$