2006-11-28 03:20:20 · 6 answers · asked by silentalarm 1 in Science & Mathematics ➔ Mathematics
want the proof? Here it is ( i am assuming you know the dervatives of sinx and cosx)
y = tanx
y = sinx/cosx
y' = [cosx(cosx) - sinx(-sinx)]/(cosx)^2 --> quotient rule
y'= [(cosx)^2 + (sinx)^2]/(cosx)^2
y' = 1/(cosx)^2
y' = (secx)^2
voila
2006-11-28 03:27:38 · answer #1 · answered by Morkeleb 3 · 3⤊ 1⤋
Differential Of Tanx
2016-09-30 10:49:53 · answer #2 · answered by cistrunk 4 · 0⤊ 0⤋
This Site Might Help You.
RE:
What is the differential of y=tanx?
2015-08-19 01:09:19 · answer #3 · answered by ? 1 · 0⤊ 0⤋
y=tanx
d(tanx)/dx=sec^x
{standard derivative}
i hope that this helps
2006-11-28 05:37:27 · answer #4 · answered by Anonymous · 1⤊ 0⤋
y = tanx
dy/dx = (secx)^2
2006-11-28 05:51:42 · answer #5 · answered by Kemmy 6 · 0⤊ 0⤋
sec^2 x
2006-11-28 03:22:19 · answer #6 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
If you mean derivative, it's sec² x.
2006-11-28 03:28:37 · answer #7 · answered by Philo 7 · 0⤊ 0⤋
sec^2xdx
2006-11-28 03:30:14 · answer #8 · answered by raj 7 · 2⤊ 0⤋