anonymous
  • anonymous
Please Help me with this Calculus Problem Show that if y=x/tan(x) than y' = cot(x) - x(csc^2x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Maybe the first step would be to change it to y = x(cos(x))/sin(x) and then take the derivative?
Mertsj
  • Mertsj
\[y'=\frac{\tan x(1)-x(\sec ^2x)}{\tan ^2x}=
anonymous
  • anonymous
...what?

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Mertsj
  • Mertsj
Well it didn't post right. I'll try it again.
anonymous
  • anonymous
lol, yea I figured so :P
Mertsj
  • Mertsj
\[y'=\frac{\tan x-x(\sec ^2x)}{\tan ^2x}\]
anonymous
  • anonymous
Ok, so is that the answer?
Mertsj
  • Mertsj
\[\frac{\frac{\sin x}{\cos x}-\frac{x}{\cos ^2x}}{\frac{\sin ^2x}{\cos ^2x}}\]
anonymous
  • anonymous
Oh, I see
Mertsj
  • Mertsj
Now multiply top and bottom of that fraction by cos^2x You will get: \[\frac{\sin x \cos x-x}{\sin ^2x}=\frac{\sin x \cos x}{\sin ^2x}-\frac{x}{\sin ^2x}\]
anonymous
  • anonymous
is that all y'?
Mertsj
  • Mertsj
Which is: \[\frac{\cos x}{\sin x}-x \csc ^2x=\cot x-x \csc ^2x=y'\]
anonymous
  • anonymous
Is that it?
anonymous
  • anonymous
@mertsj Im confused so what do I write?
Mertsj
  • 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.

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