1. ## Implicit Differentiation

Find dy/dx by implicit differentiation.

$tan(x-y) = \frac {y}{(3 + x^2)}$

My Steps:

$\frac {d}{dx} tan(x-y) = sec^2(x-y)(1-y')$

$\frac {d}{dx} \frac {y}{(3 + x^2)} = \frac {(3+x^2)y' - y(2x)}{(3+x^2)^2}$

Therefore:

$sec^2(x-y)(1-y') = \frac {(3+x^2)y' - y(2x)}{(3+x^2)^2}$

Now I know that I need to solve for y' but no matter how long I stare at this problem I can't figure it out.

2. Originally Posted by topher0805
Find dy/dx by implicit differentiation.

$tan(x-y) = \frac {y}{(3 + x^2)}$

My Steps:

$\frac {d}{dx} tan(x-y) = sec^2(x-y)(1-y')$

$\frac {d}{dx} \frac {y}{(3 + x^2)} = \frac {(3+x^2)y' - y(2x)}{(3+x^2)^2}$

Therefore:

$sec^2(x-y)(1-y') = \frac {(3+x^2)y' - y(2x)}{(3+x^2)^2}$

Now I know that I need to solve for y' but no matter how long I stare at this problem I can't figure it out.
just do the proper grouping and distribution..

$(1-y')\sec^2(x-y) = \sec^2{(x-y)} - y'\sec^2{(x-y)} = \frac {(3+x^2)y' - y(2x)}{(3+x^2)^2}$

$(3+x^2)^2\sec^2{(x-y)} - (3+x^2)^2y'\sec^2{(x-y)} = (3+x^2)y' - y(2x)$.. how about that, can you do the isolation of $y'$? Ü

3. $(3+x^2)^2\sec^2{(x-y)} - (3+x^2)^2y'\sec^2{(x-y)} = (3+x^2)y' - y(2x)$

Would then become:

$(3+x^2)^2\sec^2{(x-y)} + y(2x) = (3+x^2)y' + (3+x^2)^2y'\sec^2{(x-y)}$

Which simplifies to:

$y'((3+x^2)+(3+x^2)^2\sec^2{(x-y)}) = (3+x^2)^2\sec^2{(x-y)} + y(2x)$

So the final answer would be:

$y' = \frac {(3+x^2)^2\sec^2{(x-y)} + y(2x)}{(3+x^2)+(3+x^2)^2\sec^2{(x-y)}}$

Is this right?

4. yes.. Ü