Question

Find the derivative of arccos(x) through first principles.

arccos(x) aka the inverse function of cos(x).
How would you go about with doing this?

Answer 1

first principles? are you not allowed to use the chain rule?

Answer 2

OH YEAH!

Answer 3

if you are allowed to use the chain rule and the fact that
\[\cos(\arccos(x))=x\] then it is straighforward. if you have to use a limit good luck

Answer 4

Wait how does that work? With the chain rule?

Answer 5

in that case take the derivative of both sides and get
\[\sin(\arccos(x))\times \arccos'(x)=1\] and then solve for
\[\arccos'(x)\]

Answer 6

I don't understand.

Answer 7

Wait, so you are starting with the property of how inverses can "undo" each other.

Answer 8

you get
\[\arccos'(x)=\frac{1}{-\sin(\arccos(x))}\]

Answer 9

right and i made a mistake

Answer 10

the derivative of cosine is -sine

Answer 11

so we start with the identity
\[\cos(\arccos(x))=x\]
then take the derivative. you get
\[-\sin(\arccos(x))\times \arccos'(x)=1\] by the chain rule (on the left)

Answer 12

Ok wait, so you are doing:
-sin(arccos(x))*arccox'(x) = 1

Answer 13

we are pretending we don't know what
\[\arccos'(x)\] is so we are going to solve for it using algebra

Answer 14

Ok proceed please.

Answer 15

first is it clear that the derivative of
\[f(g(x))\] is
\[f'(g(x))\times g'(x)\]?

Answer 16

because that is what i used on the left hand side

Answer 17

Yes so far.

Answer 18

ok and in fact there is nothing special about cosine and arccosine. we could have just as well used f and f inverse

Answer 19

so we are at this line
\[-\sin(\arccos(x))\times \arccos'(x)=1\]

Answer 20

i solve for
\[\arccos'(x)\] in one simple algebra step and get
\[\arccos'(x)=\frac{1}{-\sin(\arccos(x))}\]

Answer 21

what the heck? Why is my solution to someone else's question up there?

Answer 22

I just opened another window to answer someone else's question, and the answer I gave them is on this question..

Answer 23

of course we are not done, but that is the idea. we now have to compute
\[-\sin(\arccos(x))\] which is some trig work, but before we do it notice that if i replace cosine by f and arccosine by f inverse we get
\[\frac{d}{dx}f^{-1}(x)=\frac{1}{f'(f^{-1}(x)}\]

Answer 24

a very general statement that says if i want the derivative of an inverse function, i can find it by taking the reciprocal of the derivative of the original function evaluated at the inverse

Answer 25

one minute. We started off with:
cos(arccos(x)) = x right?

Answer 26

again wrong!
\[-\sin(\arccos(x))=-\sqrt{1-x^2}\]

Answer 27

Who are you talking to?

Answer 28

yes we started there. then take the derivative of both sides

Answer 29

OHH Ok I see.

Answer 30

the derivative of x is 1

Answer 31

Got it. Ok then countinue from 1/-sin(arccos(x) = arccos'(x)

Answer 32

the derivative of
\[\cos(\arccos(x))\] is
\[-\sin(\arccos(x))\times \arccos'(x)\] via the chain rule

Answer 33

Yes

Answer 34

oh no!

Answer 35

hold on

Answer 36

so we have
arccos'(x) = 1/-sin(arccos(x))

Answer 37

right that line is correct

Answer 38

now we need to find an algebraic expression for
\[-\sin(\arccos(x))\] do you know how to get that one?

Answer 39

no

Answer 40

ok lets call arcos(x) say theta, the angle whose cosine is x. then make a triangle

Answer 41

that ugly theta is supposed to represent arccos(x), the angle whose cosine is x

Answer 42

that is why i labeled the "adjacent" side x and they hypotenuse 1

Answer 43

because x / 1 = x

Answer 44

now we want the sine of that angle, which amounts to finding the other side, which we do in one step by pythagoras. it is
\[\sqrt{1-x^2}\]