Question

How do you find `(dy)/(dx)` given `xcos(2x+3y)=ysinx`?

Answer 1

`dy/dx = (cos(2x+3y)-2xsin(2x+3y) - ycos x)/(sinx +3xsin(2x+3y))`

Explanation:

We have:

`xcos(2x+3y) = ysin x`

Method 1 - Implicit Differentiation

Applying the product rule and chain rule we get:

`(x)(d/dx cos(2x+3y))+(d/dx x )(cos(2x+3y)) = (y)(d/dxsin x)+(d/dx y)(sin x)`

`:. x(d/dx cos(2x+3y))+cos(2x+3y) = ycos x+sinx dy/dx`

`:. x(-sin(2x+3y)d/dx (2x+3y))+cos(2x+3y) = ycos x+sinx dy/dx`

`:. -xsin(2x+3y) (2+3dy/dx)+cos(2x+3y) = ycos x+sinx dy/dx`

Now we multiply out and collect terms:

`:. -2xsin(2x+3y) -3xsin(2x+3y)dy/dx+cos(2x+3y) = ycos x+sinx dy/dx`

Factoring out `dy/dx` we get:

`:. dy/dx(sinx + 3xsin(2x+3y)) = cos(2x+3y)-2xsin(2x+3y) - ycos x`

`:. dy/dx = (cos(2x+3y)-2xsin(2x+3y) - ycos x)/(sinx +3xsin(2x+3y))`

Method 2 - Using the Implicit Function Theorem:

`dy/dx = -((partial f)/dx)/((partial f)/dy)`

Where:

`f(x,y) = 0`

We have:
Let:

`f(x,y) = xcos(2x+3y) - ysin x`

Then the partial derivatives are:

# (partial f)/(partial x) = (x)((partial )/(partial x) cos(2x+3y))+ ((partial )/(partial x)x)(cos(2x+3y)) - ycos x#

`\ \ \ \ \ \ \ = x( -2sin(2x+3y))+cos(2x+3y) - ycos x`

`\ \ \ \ \ \ \ = -2xsin(2x+3y)+cos(2x+3y) - ycos x`

And:

`(partial f)/(partial y) = -xsin(2x+3y)(3) - sin x`
`\ \ \ \ \ \ \ = -(3xsin(2x+3y) + sin x)`

And so:

`dy/dx = -(-2xsin(2x+3y)+cos(2x+3y) - ycos x)/(-(3xsin(2x+3y) + sin x))`
`\ \ \ \ \ \ = (-2xsin(2x+3y)+cos(2x+3y) - ycos x)/(3xsin(2x+3y) + sin x)`