JoBo
Please Help me with this Calculus Problem
Show that if y=x/tan(x) than y' = cot(x) - x(csc^2x)
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brinethery
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Maybe the first step would be to change it to y = x(cos(x))/sin(x) and then take the derivative?
Mertsj
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\[y'=\frac{\tan x(1)-x(\sec ^2x)}{\tan ^2x}=
JoBo
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...what?
Mertsj
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Well it didn't post right. I'll try it again.
JoBo
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lol, yea I figured so :P
Mertsj
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\[y'=\frac{\tan x-x(\sec ^2x)}{\tan ^2x}\]
JoBo
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Ok, so is that the answer?
Mertsj
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\[\frac{\frac{\sin x}{\cos x}-\frac{x}{\cos ^2x}}{\frac{\sin ^2x}{\cos ^2x}}\]
JoBo
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Oh, I see
Mertsj
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Now multiply top and bottom of that fraction by cos^2x
You will get:
\[\frac{\sin x \cos x-x}{\sin ^2x}=\frac{\sin x \cos x}{\sin ^2x}-\frac{x}{\sin ^2x}\]
JoBo
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is that all y'?
Mertsj
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Which is:
\[\frac{\cos x}{\sin x}-x \csc ^2x=\cot x-x \csc ^2x=y'\]
JoBo
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Is that it?
JoBo
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@mertsj Im confused so what do I write?
Mertsj
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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.