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)
2007-01-02 17:01:51
·
answer #1
·
answered by Darkness1089 2
·
0⤊
0⤋
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.
2007-01-02 17:29:34
·
answer #2
·
answered by MathBioMajor 7
·
0⤊
0⤋
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. ;-)
2007-01-02 17:02:49
·
answer #3
·
answered by Scarlet Manuka 7
·
1⤊
0⤋
[1/(1+(2x)^2)]*2
Because d/dx (arctan(x)) = 1/(1+x^2) combined with the chain rule.
2007-01-02 17:07:52
·
answer #4
·
answered by Anonymous
·
0⤊
0⤋
Use chain rule.
dy/dx = [dy/d(2x)][d(2x)/dx] = [1/(1+(2x)^2][2] = 2/(1+4x^2)
2007-01-02 17:16:50
·
answer #5
·
answered by sahsjing 7
·
0⤊
0⤋
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)]
2016-12-15 14:27:05
·
answer #6
·
answered by hayakawa 4
·
0⤊
0⤋