How do you find dy/dx of a function when x is the subject

Finding dy/dx of a function when x is the subject...?

I have the equation here but I want to solve it myself so I will give a different example...

Find dy/dx of x=sin2y

one trick you can use is to find dx/dy

in your given equation dx/dy = 2cos2y

now we know that dy/dx *dx/dy = 1

this is true because

dy/dy *dx/dx = 1*1 = 1


now this means that dy/dx = 1/(dx/dy)

so you found dx/dy

so dy/dx = 1/(2cos2y)

its not always neat but it does work

you can use implicit differentiation

x = sin2y

Take derivative of both sides. On the LHS you will have d/dx (x) which is 1. and on the RHS you can use chain rule:

d/dx(x) = d/dx(sin2y)

1 = cos2y * [d/dx (2y)]

1 = cos2y * 2 dy/dx

dy/dx = 1/(2cos2y)

If you want to write it in terms of x,
dy/dx = 1/[2√(1-x²)]

(because cos²(2y) = 1-sin²(2y) )

or you can rewrite your equation as

2y = sinˉ¹(x)
y = (1/2)sinˉ¹(x)
y ' = 1/[2√(1-x²)]

You differentiate implicitly. I won't go into much detail as to why it works, but if you are interested I suggest reading the chapter on your calculus book.

dy/dx of x = sin2y

1 = 2cos(2y)(dy/dx)
dy/dx = (1/2)cos(2y)

I'd solve for y:

y = (1/2) arcsin(x) + n*pi

and then take dy/dx:

dy/dx [(1/2) arcsin(x) + n*pi] = 0.5/(square root(1- x^2))

nothin to do hard wrk......simply fine dx/dy thn inverse ur answer u will get dy/dx....esy wayyy

you treat x as a constant therefore dy/dx of x is just 0.