Question

use implicit differentiation to find the derivative of

Answer 1

\[f(x) = \cos^{-1}(x)\]

Answer 2

y = cos-1(x) means

cos(y) = x ; now just dwrtx

Answer 3

y'.-sin(y) = 1

y' = 1/-sin(y)

Answer 4

sin(y) just so happens to be: sin(cos-1(x))

Answer 5

put
\[f(x)=\cos(x)\]
\[f(f^{-1}(x))=x\]
take the derivative using chain rule (or implicitly if you like)

\[f'(f{-1}(x))\times( f^{-1}(x))'=1\]

Answer 6

making
\[(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}\]

Answer 7

previous formula is true for any f and f inverse. so if you replace f by cosine you get
\[(\cos^{-1})'=\frac{1}{\sin(cos^{-1}(x))}\]

Answer 8

hello amistre!

Answer 9

cos-1(x) means that some ratio of x/1 is the cosine of an angle; this is where a picture comes in handy but;

cos(a = x/1; such that sin(a) = (B)/1 = sqrt(1-x^2)

| \
| \ (1)
(B) | \
| a \
|----------
(x)

Answer 10

howdy sat ;)

Answer 11

i will be quite and let amistre show you that
\[\sin(\cos{-1}(x))=\sqrt{1-x^2}\]

Answer 12

nice triangle!

Answer 13

lol .... it was all i had :)

Answer 14

the other inverses go by the same logic

Answer 15

jeez i cannot type today

Answer 16

\[\sin(cos^{-1}(x))=\sqrt{1-x^2}\]

Answer 17

btw i think we both made a minor error yes?

Answer 18

oh no, only i did not you

Answer 19

it is
\[\frac{d}{dx} \cos^{-1}(x)\]=\[\frac{1}{-\sin(\cos^{-1}(x))}\]=
\[-\frac{1}{\sqrt{1-x^2}}\]

Answer 20

Thanks, makes sense. I need to revise implicit differentiation a little more

Answer 21

implicit and explicit are the same thing .... dont let anyone tell you otherwise