
Victoria V. answered 04/11/16
Tutor
5.0
(402)
Math Teacher: 20 Yrs Teaching/Tutoring CALC 1, PRECALC, ALG 2, TRIG
If you need to do this by the definition involving limits, then this would be one way to go about it:
The derivative by limits definition is:
lim f(x+h) - f(x)
h -->0 ---------------
h
So
f(x) = tan x = sin x
--------
cos x
and
f(x+h) = tan (x+h) = sin(x+h)
----------
cos(x+h)
Putting these into the definition:
lim sin(x+h) - sin x
h-->0 ----------- ------
cos (x+h) cos x
-----------------------------------
h
Need a common denominator for top subtraction, so multiply left fraction by (cosx/cosx) and multiply right fraction by cos(x+h)/cos(x+h). This results in
lim sin(x+h) cos(x) - sin(x) cos(x+h)
h-->0 --------------------------------------------
cos(x+h) cos(x)
------------------------------------------------------
h
Now "fix" the complex fraction by multiplying the top fraction by [cos(x+h) cos(x)] to get rid of the denom in the top, and multiply the "h" in the bottom by [cos(x+h) cos(x)] so that you have really just multiplied the whole fraction by 1.
lim sin(x+h) cos(x) - sin(x) cos(x+h)
h-->0 ------------------------------------------
h cos(x+h) cos(x)
Right here, we need a trig identity.
sin(A-B) = sin(A)cos(B) - cos(A)sin(B)
This is our situation if A = x+h and B = x, so can rewrite the numerator as sin((x+h) - x) and this is sin(h)
Putting it all together:
lim sin(h)
h-->0 ---------------------------
h cos(x+h) cos(x)
Grouping this very carefully, we can use the fact that
lim sin(h)
h-->0 ---------- = 1
h
So
lim sin(h) 1 1
h-->0 --------- . ------ . --------
h cos(x+h) cos(x)
= 1 . 1 . 1
--------- --------
cos(x) cos(x)
= 1 . sec(x) . sec(x) = sec2x