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Trig in Calculus
Overview
Well, you pretty much should have all the derivatives down
pat by now. The only thing I haven’t gone into is trigonometric functions. I
figured I’d make everything complicated after I’m done with the formulas.
Incidentally, if you don't feel comfortable with the material yet, I have
problem sets linked at the end of every section. I don't know if I've made that
clear.
The derivates of all trig functions are in the reference
table. However, I am just going to work with 3 basic trig derivatives here.
y = sin x
y’ = cos x
y = cos x
y’ = -sin x
y = tan x
y’ = sec2 x
These are rules that you will have to follow. They are also
very basic, and can be complexified in interesting ways. (Did I just create a
word?)
In the spirit of having you actually remember some of the above, I will allow
you to see how the derivative of sin x is cos x. Carefully draw both a sin and
cos curve on one graph, from 0 to 2∏.
(That symbol is pi) You will notice at x = 0 that the angle that the sin curve
goes through the origin is a 45 degree slop upwards, which means that it goes up
1 y for every 1 x. The slope is then 1, and the derivative at that point should
be 1. Cos of 0 is 1. At x = ∏/2, sin x reaches a maximum, at which the
slope is 0. Cos ∏/2 is 0. At ∏, the sin curve has a -1 slope, and
cos is -1. This pattern is correct not only at these points but at every point.
A similar analysis will show that the derivative of cos x is -sin x.
With complex rules
y = 5sin x
y’ = 5cos x This is
because the 5 is a coefficient, and any time you are taking a derivative of a
function with a number in front, you can take that number away for the purposes
of taking the derivative, and put it back
in after solving.
y = sin 5x
y’ = 5cos 5x This is
simply the chain rule, used in the last topic. The 5x is u and the derivative of
sin u is cos u. The derivative of 5x is 5. Multiply the derivative of outside by
derivative of inside, and you have the derivative of equation. Then you replace
the u.
y = 3sin 5x
y’ = 15cos 5x
y = 3cos 5x
y’ = -15sin 5x
y = sin2 x = (sin
x)2 = u2 (and u = sin x)
y’ = 2u(cos x) = 2sin x cos x
This one I solved by a chain rule similar to what you saw in the chain
rule section. Notice that I put the square around the entire sin x, this is
explained in the Intro. Now I made sin x the inner function, and u2
the outer function. The rest is simple. Take the derivatives of both and
multiply. Then replace the u to its original status (sin x).
y = 3sin2 (5x + 3) This one is not as tough as it
looks if you understood everything before. On the other hand, if you understood
everything before it probably doesn’t look too tough. This requires a chain
rule inside a chain rule. The equation rearranges as
3[sin (5x + 3)]2
The outside equation to solve is u2 and u = sin
(5x +3). So If I take the derivative of the outside, getting 2u, and multiply it
by the derivative of the inside function, then I have my answer. But what is the
derivative of the inside function? The derivative of sin (5x +3) requires a
chain rule. Once you figure out the derivative, take it back and multiply it by
the other one. The derivative of this is 5 * cos (5x + 3). Multiply that by 2u,
and multiply by the 3 that we've been ignoring, and you have 30u cos (5x + 3) which is
30[sin (5x + 3)]cos (5x + 3). I hope I
didn’t screw that up. No, I think it’s right.
y = 2tan4 x
y’ = 8tan3 x sec2 x This is nothing you can’t handle. The u turned back
into a tan x at the end, and 8u3sec2x became 8tan3
x sec2 x.
y = sin (ln x)
y’ = cos (ln x) * 1/x
y = ln (sin x)
y' = (1/sin x) * (cos x) = cos x/sin x = cot x
y = ln (cos x)
y' = (1/cos x) * (- sin x) = - sinx/cos x = - tan x
y = cos (ln 2x)
y’ = -sin (ln 2x) * 1/x I
used two chains in that one. The 2x must become a u in order to get the
derivative of a ln 2x.
The following example uses a derivative that you are required to know, and is
shown along with the rest in my reference tables.
y = tan x/sin x = sec
x y' = sec x tan x
y = tan (exx2)
y’ = sec2 (exx2) * (exx2
+ ex2x) That one has a product rule inside the chain rule. Make
sure your sec2 is left with a (exx2) inside,
because during the solving process it turns into a u.
OK, I think I’ve done enough trigonometries for one day.
I’ll move on to implicit differentiation.
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