Question

y=xcosy+ycosx then find dy/dx
​

Answer 1

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Differentiation by Chain rule and Product rule.

Now basically here we are going to use the Chain rule and the product rule.

since here,

y = xcosy + ycosx

then we get as,

dy/dx = x. (-sinydy/dx) + cosy. (1) + y. ( -sinx) + cosx. (dy/dx)

now as we obtained this expression by applying the chain rule ane product rule,

We have to seperate the dy/dx terms at one side.

dy/dx = -xsiny. dy/dx + cosx. dy/dx + cosy -y. sinx

dy/dx + xsiny. dy/dx - cosx. dy/dx = cosy - ysinx

dy/dx ( 1 + xsiny - cosx) = cosy - ysinx

so we finally get as,

⭐dy/dx = (cosy -ysinx) ÷ (1+xsiny-cosx)

Answer 2

Here is the answer different than given above.

Using the properties of partial differentiation.

Let t = x cos y + y cos x- y

now, ∂t/∂y = - x sin y + cos x -1

then, dy/ dx = - (∂t/∂x) / (∂t/∂y) = - ( cos y - y sin x) / (-x sin y + cos x - 1)

or, dy / dx = (cos y - y sin x ) / ( 1 + x sin y - cos x)