dx/dy?!?!?

Question

how on earth will i get dx/dy (derivative of x with respect to y... is that right?) of any function? for example: sec^2(x+y^2)=(2x+y)^25

is the process the same with dy/dx? are you just going to look for dx/dy instead of dy/dx using implicit differentiation?

is dx/dy=1/f'(x)? so to get dx/dy, i'm going to get f'(x) first then reciprocate?

Answer 1

Both ways are correct. You can take y as function of x and derive implicitly for x to find dy/dx. The reciprocal of the result is dx/dy.
The direct way is to treat x as function of y an derive for y.

Let f(x,y) = 0

Take y is a function of x and derive for x
df/dx = ∂f/∂x + ∂f/∂y × dy/dx = 0
=>
dy/dx = -(∂f/∂x) / (∂f/∂y)
=>
dx/dy = -(∂f/∂y) / (∂f/∂x)

If you treat x as function of y and derive for y
df/dy = ∂f/∂x × dx/dy + ∂f/∂y = 0
=>
dx/dy = -(∂f/∂y) / (∂f/∂x)
which is surprisingly the same as above.

For the given expression
dy/dx = [4ysec²( x+y²)tan( x+y²) -25(2x+y)^24] / [2sec²( x+y²)tan( x+y²) -50(2x +y)^24]

Answer 2

The process is the same with dy/dx, except you differentiate with respect to y, and for every instance of x, differentiating it would be dx/dy.

sec^2(x + y^2) = (2x + y)^25

Differentiating with respect to y,

2sec(x + y^2) [dx/dy + 2y] = 25(2x + y)^(24) [2(dx/dy) + 1]

Answer 3

Example
x = y³ + 8y² + 3y + 2
dx/dy = 3y² + 16y + 3
ie similar method to that for finding dy/dx only this time are differentiating with respect to y.

The equation at the start of your question looks dodgy to me-----error in typing?

Answer 4

dx/dy is actually the reciprocal of dy/dx

Answer 5

Yep, just flip the way you've been thinking. You're in a rut thinking about x's and y's.

Answer 6

Pretty much, just treat x as y and vice versa.