Question

find dy/dx from x sin (xy) + cos(xy)=0​

Answer 1

Answer:

dy/dx = - (x²y + y - 1) / {x (x² + 1)}

Step-by-step explanation:

We know that

d/dx (uv) = u dv/dx + v du/dx,

where u, v are functions of x.

Given,

x sin(xy) + cos(xy) = 0

or, x sin(xy) = - cos(xy)

or, x tan(xy) = - 1

or, x tan(xy) + 1 = 0 ... (1)

Now differentiating both sides with respect to x, we get

d/dx {x tan(xy) + 1} = 0

or, x d/dx {tan(xy)} + tan(xy) dx/dx = 0

or, x sec²(xy) d/dx (xy) + tan(xy) = 0

or, x sec²(xy) (x dy/dx + y) + tan(xy) = 0

or, x² sec²(xy) dy/dx + xy sec²(xy) + tan (xy) = 0

or, x² sec²(xy) dy/dx = - {xy sec²(xy) + tan (xy)}

or, dy/dx = - {xy sec²(xy) + tan (xy)} / {x² sec²(xy)}

= - [xy {1 + tan²(xy)} + tan(xy)] / [x² {1 + tan²(xy)}]

= - [xy {1 + (- 1/x)²} + (- 1/x)] / [x² {1 + (- 1/x)²}] , by (1)

= - [xy (1 + 1/x²) - 1/x] / [x² (1 + 1/x²)]

= - (xy + y/x - 1/x) / (x² + 1)

= - (x²y + y - 1) / {x (x² + 1)} ,

which is the required derivative.