Question

find dy/dx when: sin² x + 2cos y + xy =0​

Answer 1

$\huge\tt \blue{answer}$

$\large\bf \pink{\frac{(y + \sin2x)}{( 2\sin \: y - x)} }$

$\large\bf \red{solution}$

Given sin² x + 2 cos y + xy = 0

Differentiate both sides w.r.t.x

2 sin x cos x – 2 sin y(dy/dx) + (y + x dy/dx) = 0

(dy/dx)( x – 2 sin y) = (-y – 2 sin x cos x)

(dy/dx)( x – 2 sin y) = (-y – sin 2x)

(dy/dx) = (-y – sin 2x)/( x – 2 sin y)

= -(y + sin 2x)/( x – 2 sin y)

= (y + sin 2x)/( 2 sin y – x)

• Brainliest plz!-,-

Answer 2

$\huge\tt \blue{answer}$

$\large\bf \pink{\frac{(y + \sin2x)}{( 2\sin \: y - x)} }$

$\large\bf \red{solution}$

Given sin² x + 2 cos y + xy = 0

Differentiate both sides w.r.t.x

2 sin x cos x – 2 sin y(dy/dx) + (y + x dy/dx) = 0

(dy/dx)( x – 2 sin y) = (-y – 2 sin x cos x)

(dy/dx)( x – 2 sin y) = (-y – sin 2x)

(dy/dx) = (-y – sin 2x)/( x – 2 sin y)

= -(y + sin 2x)/( x – 2 sin y)

= (y + sin 2x)/( 2 sin y – x)

Brainliest plz!-,-