prove it to me that it equals to tanx
2007-10-03 15:26:31 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
It does not:
d/dx (sec x) = sec(x)tan(x)
y = 1/cos(x) = [cos(x)]^(-1)
dy/dx = -1*[cos(x)]^(-2) * (-sin(x))
= sin(x) / cos²(x)
= [1/cos(x)][sin(x)/cos(x)]
= sec(x)tan(x)
2007-10-03 15:29:50 · answer #1 · answered by gudspeling 7 · 2⤊ 2⤋
As mentioned, it is equal to sec x tan x. This can be seen using the power and chain rules, as follows:
d/dx (sec x)
d/dx ((cos x)^(-1))
-1 (cos x)^(-2) * d/dx (cos x)
-1/cos² x * (-sin x)
sin x/cos² x
1/cos x * sin x/cos x
sec x tan x
2007-10-03 22:33:02 · answer #2 · answered by Pascal 7 · 0⤊ 3⤋
you can also rewrite this as 1/cos(x) and use the quotient rule from that. Then it will lead you to the answer with a few identities to work with.
2007-10-03 22:32:56 · answer #3 · answered by PinoyPlaya 3 · 0⤊ 1⤋
It's not tan x, it is sec(x) * tan(x)
As for proof, convert sec(x) to 1/cos(x)
Use the quotient rule.
2007-10-03 22:31:14 · answer #4 · answered by z_o_r_r_o 6 · 0⤊ 1⤋
well that is going to be hard since it equals sec(x)tan(x)
2007-10-03 22:29:58 · answer #5 · answered by Mαtt 6 · 0⤊ 1⤋
f (x) = sec x = 1 / cos x
f `(x) = [cos x (0) - 1 (- sin x) ] / ( cos ² x )
f `(x) = sin x / ( cos ² x )
f `(x) = tan x sec x
2007-10-04 04:52:48 · answer #6 · answered by Como 7 · 2⤊ 3⤋