Derivative of y=arctan(2x) ?

Answer 1

sub 2x as u

and dy/dx = 1 / (1 + u^2) * du/dt

and du/dt = 2

therefore:

dy/dx = ( 1 / (1 + 4x^2) ) * 2

dy/dx = 2 / (1 + 4x^2)

Answer 2

Let y = arctan (2x).

Then 2x = tan y.

We can differentiate both sides directly to get:

2 dx = sec^2 (y) dy.

2 / sec^2 (y) = dy / dx.

Recall that sec^2 (y) = 1 + tan^2 (y). Then back substitute it into the preceding equation.

2 / sec^2 (y) = 2 / [1 + tan^2 (y)] = 2 / [1 + (2x)^2] = 2 / (1 + 4x^2) = dy / dx.

So, 2 / (1 + 4x^2) = dy / dx.

Answer 3

tan y = 2x: implicit differentiation w.r.t. x gives
sec^2 y dy/dx = 2
=> (1 + tan^2 y) dy/dx = 2
=> dy/dx = 2 / (1 + (2x)^2) = 2 / (1 + 4x^2).

You can get this more simply if you know that d/dx arctan x = 1 / (1+x^2), but I find it easier to remember how to derive it. ;-)

Answer 4

[1/(1+(2x)^2)]*2

Because d/dx (arctan(x)) = 1/(1+x^2) combined with the chain rule.

Answer 5

Use chain rule.

dy/dx = [dy/d(2x)][d(2x)/dx] = [1/(1+(2x)^2][2] = 2/(1+4x^2)

Answer 6

prepare the quotient rule: y' = [-sin(x) * (a million + sin(x)) - cos(x) * cos(x)] / [a million + sin(x)]^2 strengthen each and all of the brackets: y' = [-sin(x) - sin^2(x) - cos^2(x)] / [a million + sin(x)]^2 y' = [-sin(x) -a million(sin^2(x) + cos^2(x))] / [a million + sin(x)]^2 y' = [-sin(x) - a million] / [a million + sin(x)]^2 y' = -a million[sin(x) + a million] / [a million + sin(x)]^2 y' = -a million / [a million + sin(x)]