Question

Help, implicit differentiation:

Answer 1

\[\LARGE tan(x-y)=\frac{y}{4+x^2}\]

Answer 2

first use this
tan(X - Y) = [ tanX - tanY ] / [ 1 + tanX tanY]

Answer 3

Here, Luigi, we ahve an implicit function; the expression you've typed in has not been solved for the variable y. But you can apply the derivative operator to both sides of this equation. We proceed on the assumption that y is a function of x; therefore, dx/dx = 1, but dy/dx is not = to 1 (most of the time, at least).

Answer 4

first, would you type in the formula for the derivative of the simple tangent function, y=tan x?

Answer 5

\[\large \frac{d}{dx}tanx=sec^2x\]

Answer 6

\[\frac{ d }{ dx }\tan \left( x-y \right)=\sec ^2\left( x-y \right)\left\{ 1-\frac{ dy }{ dx } \right\}\]
\[=\left( 1+\tan ^2\left( x-y \right) \right)\left\{ 1-\frac{ dy }{ dx } \right\}\]

Answer 7

Here surjithayer is applying your formula for the derivative of tan x, but he has also recognized that (x-y) is itself a function, and so he applied the chain Rule. Are you familiar with that, Luigi?

Answer 8

As before, we assume that y is a function of x, but that x is the independent variable.
Are you able to diffferentiate with respect to x: (x-y)? See what surji has done.

Answer 9

Yea, but is it necessary to change the trig function to an identity?

Answer 10

No; I wouldn't do that myself.
When you give me the go-ahead, we'll now look at the right side of the equation you're differentiating.

Answer 11

What is\[\frac{ d }{ dy }\frac{ y }{ 4+x^2 }?\]

Answer 12

Okay, so I differentiated it this way:
\[\LARGE sec^2(x+y)*(1+y')=\frac{(4+x^2)(y')-(2x)(y)}{(4+x^2)^2}\]

Answer 13

Really, really nice! Could you put into words the next step that you must carry out?

Answer 14

I have to simplify both sides if I'm not mistaken.. then get all terms with y' to one side and the rest on the other side.

Answer 15

:U

Answer 16

How am I suppose to isolate the y' from here?

Answer 17

Tagged me?

Answer 18

Yes!