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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
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

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brinetheryBest ResponseYou've already chosen the best response.0
Maybe the first step would be to change it to y = x(cos(x))/sin(x) and then take the derivative?
 2 years ago

MertsjBest ResponseYou've already chosen the best response.1
\[y'=\frac{\tan x(1)x(\sec ^2x)}{\tan ^2x}=
 2 years ago

MertsjBest ResponseYou've already chosen the best response.1
Well it didn't post right. I'll try it again.
 2 years ago

MertsjBest ResponseYou've already chosen the best response.1
\[y'=\frac{\tan xx(\sec ^2x)}{\tan ^2x}\]
 2 years ago

JoBoBest ResponseYou've already chosen the best response.0
Ok, so is that the answer?
 2 years ago

MertsjBest ResponseYou've already chosen the best response.1
\[\frac{\frac{\sin x}{\cos x}\frac{x}{\cos ^2x}}{\frac{\sin ^2x}{\cos ^2x}}\]
 2 years ago

MertsjBest ResponseYou've already chosen the best response.1
Now multiply top and bottom of that fraction by cos^2x You will get: \[\frac{\sin x \cos xx}{\sin ^2x}=\frac{\sin x \cos x}{\sin ^2x}\frac{x}{\sin ^2x}\]
 2 years ago

MertsjBest ResponseYou've already chosen the best response.1
Which is: \[\frac{\cos x}{\sin x}x \csc ^2x=\cot xx \csc ^2x=y'\]
 2 years ago

JoBoBest ResponseYou've already chosen the best response.0
@mertsj Im confused so what do I write?
 2 years ago

MertsjBest ResponseYou've already chosen the best response.1
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.
 2 years ago
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