Question

derivative of inverse function
z=cos^3 2x
y=cot^2 1/4x
y=tan 2x/1-cot 2x
pls. show your sloutions..pls

Answer 1

you want the derivative of
\[\cos^3(2x)\]?

Answer 2

or maybe i am confused because it said "inverse function.

Answer 3

in any case the derivative of
\[\cos^3(2x)\] is a chain rule problem. the derivative of something cubed is three something squared. the derivative of cosine of something is minus sine of something and the derivative of 2x is 2, so you get
\[3\cos^2(2x)\times (-\sin(2x))\times 2= -6\cos^2(2x)sin(2x)\]

Answer 4

am i want the derivative of inverse function....

Answer 5

what inverse function?

Answer 6

example y=f(x)
y^-1=f^-1(x)
y=3x+2
Find the inverse
y^-1
3x=y-2
x=y-2/3
y^-1= x-2/3

Answer 7

do you mean you want the derivative of
\[f(x)=\frac{1}{2}\cos^{-1}(\sqrt[3]{x})\]?

Answer 8

oh you want to find the inverse function. not the "derivative' as in calculus. ok i found the first one but i didn't write the steps. i can if you like

Answer 9

can you pls...

Answer 10

start with
\[x=\cos^3(2y)\]
take the cube root get
\[\sqrt[3]{x}=\cos(2x)\] take the inverse cosine get
\[\cos^{-1}(\sqrt[3]{x})=2y\] divide by 2 get
\[f^{-1}(x)=\frac{1}{2}\cos^{-1}(\sqrt[3]{x})\]

Answer 11

now do you need the "derivative" of this thing? for a calculus class?

Answer 12

yes

Answer 13

ok well now we have a chain rule problem. you can ignore the 1/2 it just sits out front

Answer 14

am excuse me.can i show you an example from our instructor?

Answer 15

the derivative of
\[\cos^{-1}(x)=-\frac{1}{\sqrt{1-x^2}}\] so the derivative of
\[\cos^{-1}(\sqrt[3]{x})=-\frac{1}{1-(\sqrt[3]{x})^2)}\times \frac{1}{3}x^{-\frac{2}{3}}\]

Answer 16

sure
my answer was wrong anyway there is a typo

Answer 17

\[\cos^{-1}(\sqrt[3]{x})=-\frac{1}{\sqrt{1-(\sqrt[3]{x})^2)}}\times \frac{1}{3}x^{-\frac{2}{3}}\]

Answer 18

maybe i am confused as the the question. go ahead and give and example