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Limit definition, derivative of sec?
I've been spending too much time trying to put it all together, could someone please help to prove (using limit definition h--->0) that d/dx (sec x) = sec x tan x.
Thanks so much, I am unable to see how to fit together all the identities using the limit definition.
1 Answer
- MathGuyLv 61 decade agoFavorite Answer
One of the identities you need is the sum of angles identity
cos(a+b) = cos(a)cos(b) - sin(a)sin(b).
You also need to know
sin(x)/x = 1 as x-->0 and
(cos(x)-1)/x = 0 as x-->0
First prove derivative of cos(x):
d/dx[cos(x)] = lim h-->0 [cos(x+h)-cos(x)]/h =
[cos(x)cos(h)-sin(x)sin(h) - cos(x)] / h =
[cos(x)[cos(h)-1]-sin(x)sin(h)] / h =
cos(x)[cos(h)-1] / h - [sin(x)sin(h)] / h =
[cos(h)-1]/h * cos(x) - sin(h)/h * sin(x) =
0 * cos(x) - 1 * sin(x) as h-->0 =
-sin(x)
d/dx(cos(x)) = -sin(x)
Now we can find d/dx[sec(x)]
d/dx(sec(x)) = d/dx(1/cos(x))
Use the quotient rule:
d/dx(sec(x)) = [0*cos(x) - 1*(-sin(x))] / [cos²(x)]
d/dx(sec(x)) = sin(x) / cos²(x)
d/dx(sec(x)) = sin(x) / cos(x) * 1/[cos(x)]
d/dx(sec(x)) = tan(x)sec(x)
