The Student Room Group
implicit differentiation (with chain and product rule):

diff wrt x:
1y + x(dy/dx) = 2e^(2x) + 2(dy/dx)e^(2y)

rearrange for dy/dx

y-2e^(2x) = 2e^(2y)(dy/dx) - x(dy/dx)
dy/dx = (y-2e^(2x))/(2e^(2y) - x)
Reply 2
Thank you very much.

I think in your solution one of the y's turned into an x.

The final answer is (by the book) (y - 2e^(2x))/(2e^(2y) - x).

Also: could you explain exactly how you got the "2(dy/dx)e^(2y)" bit in your first line.

I know it is differentiating implicitly and you have to do something with ln but I am not sure how it goes.
Ah yes sorry, will edit post!

As long as you understood the applications of implicit, chain and product rules, which is the most important thing :wink:
samdavyson
Thank you very much.

I think in your solution one of the y's turned into an x.

The final answer is (by the book) (y - 2e^(2x))/(2e^(2y) - x).

Also: could you explain exactly how you got the "2(dy/dx)e^(2y)" bit in your first line.

I know it is differentiating implicitly and you have to do something with ln but I am not sure how it goes.

let v = e^(2y)
v = e^u
dv/du = e^u
u = 2y
du/dx = 2(dy/dx)

chain rule:
dv/dx = (dv/du)(du/dx)
dv/dx = e^u . 2(dy/dx)
dv/dx = 2e^(2y) (dy/dx)
Reply 5
endeavour
let v = e^(2y)
v = e^u
dv/du = e^u
u = 2y
du/dx = 2(dy/dx)

chain rule:
dv/dx = (dv/du)(du/dx)
dv/dx = e^u . 2(dy/dx)
dv/dx = 2e^(2y) (dy/dx)


Bloody hell!

You did all of that in one step! I have got some practice to do.

Thank you very much, I am getting the general idea: if it looks hard substitute.