Question

Please Help me with this Calculus Problem



Show that if y=x/tan(x) than y' = cot(x) - x(csc^2x)

Answer 1

Maybe the first step would be to change it to y = x(cos(x))/sin(x) and then take the derivative?

Answer 2

\[y'=\frac{\tan x(1)-x(\sec ^2x)}{\tan ^2x}=

Answer 3

...what?

Answer 4

Well it didn't post right. I'll try it again.

Answer 5

lol, yea I figured so :P

Answer 6

\[y'=\frac{\tan x-x(\sec ^2x)}{\tan ^2x}\]

Answer 7

Ok, so is that the answer?

Answer 8

\[\frac{\frac{\sin x}{\cos x}-\frac{x}{\cos ^2x}}{\frac{\sin ^2x}{\cos ^2x}}\]

Answer 9

Oh, I see

Answer 10

Now multiply top and bottom of that fraction by cos^2x
You will get:
\[\frac{\sin x \cos x-x}{\sin ^2x}=\frac{\sin x \cos x}{\sin ^2x}-\frac{x}{\sin ^2x}\]

Answer 11

is that all y'?

Answer 12

Which is:
\[\frac{\cos x}{\sin x}-x \csc ^2x=\cot x-x \csc ^2x=y'\]

Answer 13

Is that it?

Answer 14

@mertsj Im confused so what do I write?

Answer 15

Your problem said to show that the first derivative is equal to cotx -xcsc^2x so start with y =x/tanx and write all those steps that I have shown.