Question

LGBADERIVATIVE:

\[\huge \tan (x-y) = \frac{y}{1+x^2}\]

Answer 1

how to start this?

Answer 2

tan(x - y) = y / (1 + x^2)

differentiating implicitly

sec^2(x - y)(1 - y ') = [ (1+x^2)y ' - 2xy ] /(1 + x^2)^2

sec^2(x - y) - y ' sec^2(x - y) = y ' / (1 + x^2) - 2xy / (1 + x^2)^2

y ' [ 1 / (1 + x^2) + sec^2(x - y) ] = 2xy / (1 + x^2)^2 + sec^2(x - y)

multiply with (1 + x^2)^2

y ' [ 1+ x^2 + (1+x^2)^2 sec^2(x - y) ] = 2xy + (1+x^2)^2 sec^2(x - y)

Answer 3

y ' =

\[2xy + (1+x^2)^2 \sec^2(x - y)\over (1 + x^2) [ 1 + (1+x^2)\sec^2(x-y)]\]

Answer 4

lol nice...

Answer 5

Is it correct @lgbasallote

Answer 6

yup i thiink it is

Answer 7

it looked intimidating at first haha guess it's just long

Answer 8

lol