



The
derivatives of trig functions are a bit different. 


Look at
the sine function, and the slope of the sine function 


at some
selected points. 





F(X)
= sin X 
X° 
slope 
0° 
1 
90° 
0 
180° 
1 
270° 
0 
360° 
1 






Look at
these angles and slope values. 


At 0°
the value is 1, 


At 90°
the value is 0, 


At 180°
the value is 1, 


At 270°
the value is 0, 


and at
360°
the value is back to 1. 





That
looks like the values of the cosine. 


And it
is: 


If
F(X) = sinX then
F'(X) = cosX






That's
like cosmically weird, but that's the way it works. 


I wonder
if the derivative of the cosine might be the sine? 


Let's
see... 





F(X)
= cos X 
X° 
slope 
sinX 
0° 
0 
0 
90° 
1 
1 
180° 
0 
0 
270° 
1 
1 
360° 
0 
0 






Well it
starts out the same as the sine at zero. 


But when
the sine goes positive, the slope of the cosine goes negative. 


And vice
a versa 


It looks
like the sine but negative ... 





BINGO! We
have a winner! 


If
F(X) = cosX then
F'(X) =  sinX






Not quite
what we were expecting, but close. 











Sometimes
we don't just have X as the variable. 


Sometimes
we have something more complicated. 


Like 2X. 


OK,
that's not so complicated. 


But it's
enough. 





If we
want to find the derivative of F(X) = sin2X, what do we do? 





It turns
out that this is one of those outside function and inside function
things. 


The outside part is G(X)
= sin(something)



And the
inside part is H(X) = 2X (the something!) 





That
means to find the derivative we use the power chain rule 


from a
couple of pages ago. 





So you
don't have to go looking, here it is: 





If
F(X) = H(G(X))
then F'(X) = H'(G(X))
x G'(X)






So the
derivative of the sine is the cosine, and the derivative of 2X is 2. 





Putting
this all together, and remembering 


that we don't touch the inside
function when we
do the outside derivative, 


we get: 





If
F(X) = sin2X then
F'(X) = cos2X ´
2



which simplifies to
F'(X) = 2cos2X












OK, 4
more trig functions to go. 


The deal is,
now we're going to cheat. 





All of the rest
of the trig functions can be written as functions of sine and
cosine. 


Then we just use the rules we already have to get the answers. 





Here's the tangent: 








So, use the quotient rule: 








So the derivative of the tangent is the secant squared. 


Let's see what
we get for the derivative of the cosecant... 











Now we have a choice here.



We can do
the derivative as a quotient rule, 


OR 


We can
rewrite this as: 


csc
X = (sin X)^{1}



And do a
power chain rule. 





We did
the quotient rule on the tangent example 


so lets
do the power chain rule on this one. 











OK, for
practice you should stop right now 


and work out the derivatives of
the other 3 trig functions. 





We'll
wait. 





Really... 





Done
already? WOW! 





There are
three suggestions I have for remembering these 6 derivatives: 





1) Don't
memorize any of it, but know what the graphs 


of sine and cosine look
like. 


Then you
can work it all out when you need it. 


Yes, this
will take a good bit of time anytime you need one. 





2) Use a
whole bunch of brute force memorization and memorize all 6. 





3)
Memorize 3, and use a trick to know the other three with no extra
effort 


(That one
sounds like a winner, read on to learn the trick) 





THE TRICK 





To learn
the derivatives of the 6 trig functions 


you actually only have to
learn 3 of them. 


The three
to learn are sine, tangent, and secant. 





IF 
THEN 
F(X)
= sin X 
F'(X)
= cos X 
F(X)
= tan X 
F'(X)
= sec^{2} X 
F(X)
= sec X 
F'(X)
= sec X tan X 






The other
three functions all start with "co" 


Cosine,
Cotangent, and Cosecant. 


Think of
the functions as having partners: 





sine 
to 
cosine 
tangent 
to 
cotangent 
secant 
to 
cosecant 






To find
the derivative of the "co" functions, 


start
with the derivatives of sine, tangent and secant, 


change
each function in the derivative to it's cofunction partner 


and put a
minus sign in front of the derivative. 





So
starting with one of the three we need to know: 





If F(x) =
secant X then F'(x) = secant X tangent X






So now to
get the derivative of the cosecant: 





If
F(X) =
cosecant
X then F'(X) =

cosecant X cotangent
X






So all 6
trig function derivatives look like this: 





IF: 
THEN: 
F(X) = sinX 
F'(X) = cosX 
F(X) = cosX 
F'(X) =  sinX 
F(X) =
tanX 
F'(X)
= sec^{2} X 
F(X) =
cotX 
F'(X)
=  csc^{2} X 
F(X) =
secX 
F'(X)
= sec X tan X 
F(X) =
cscX 
F'(X)
= csc X cot X 






copyright 2008 Bruce Kirkpatrick 
