## anonymous 4 years ago Please Help me with this Calculus Problem Show that if y=x/tan(x) than y' = cot(x) - x(csc^2x)

1. anonymous

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

2. Mertsj

$y'=\frac{\tan x(1)-x(\sec ^2x)}{\tan ^2x}= 3. anonymous ...what? 4. Mertsj Well it didn't post right. I'll try it again. 5. anonymous lol, yea I figured so :P 6. Mertsj \[y'=\frac{\tan x-x(\sec ^2x)}{\tan ^2x}$

7. anonymous

Ok, so is that the answer?

8. Mertsj

$\frac{\frac{\sin x}{\cos x}-\frac{x}{\cos ^2x}}{\frac{\sin ^2x}{\cos ^2x}}$

9. anonymous

Oh, I see

10. Mertsj

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

11. anonymous

is that all y'?

12. Mertsj

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

13. anonymous

Is that it?

14. anonymous

@mertsj Im confused so what do I write?

15. Mertsj

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.