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

1. brinethery Group Title

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

2. Mertsj Group Title

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

7. JoBo Group Title

Ok, so is that the answer?

8. Mertsj Group Title

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

9. JoBo Group Title

Oh, I see

10. Mertsj Group Title

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. JoBo Group Title

is that all y'?

12. Mertsj Group Title

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

13. JoBo Group Title

Is that it?

14. JoBo Group Title

@mertsj Im confused so what do I write?

15. Mertsj Group Title

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.