Question

y =
cosy+x(sin(xy)). by differenciation​

Answer 1

⭐《ANSWER》

$\huge\mathfrak{\underline {\underline\pink {ANSWER}}}$

↪Actually welcome to the concept of the DEFFRENTIATION BY CHAIN RULE AND MULTIPLICATION RULE

↪Now basically here , we are going to use the Chain rule ,that is given as ,

↪dy/dx = dy/dt * dt/dx

↪so applying here , we get as ,

↪y = Cosy + x (Sin (xy))

↪so now here ,

↪dy/dx = d/dx ( Cosy) + d/dx ( x (Sin (xy))

↪==》dy/dx = -Siny * dy/dx + x d/dx Sin ( xy) + Sin (xy) d/dx (x)

↪==》 dy/dx = -Siny*dy/dx + x ( Cos (xy) *( x * dy/dx + y * dx/dx ) ) + Sin ( xy) *1

↪===》 dy/dx = -Siny * dy/dx + x ( Cos (xy) *(x dy/dx +y )) + Sin (xy)

↪so now we get as ,

↪==》 dy/dx = -Siny dy/dx + x Cosy * x^2 dy/dx + xy + Sin (xy)

↪taking dy/dx as common , we get

↪==》 dy/dx + Siny dy/dx - x Cosy x^2 dy/dx = xy + Sin ( xy)

↪==》 dy/dx ( 1 + Siny - x Cosy x^2 ) = xy +Sin(xy)

↪so finally we get as,

〽⭐==》 dy/dx = (xy + Sin (xy) ) / 1 + Siny - x ^3 Cos y )