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find dy/dx y=sin(x)+(1/2)cot(x) don't give me the answer. i want to work through with it
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Pretty straight forward
What have you did?
[sin (x)]((1/2)cot(x))+(sin (x))[1/2)cot(x)]
(cos(x))(1/2)cot(x))+(sin (x))((1/2)csc^2(x))
what do i do after that?
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That is wrong
What is the dy/dx of sin, and the dy/dx of cot
sin(x) is cos(x) and cot(x) is csc^2(x) right?
First u divided 2 part dy/dx or sin = cos(x) right, then next part u take dx/dx or 1/2cot(x)
-csc^2
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So the answer is be y= cos (x) - csc^2 (x) I guess
u mean Y= cos (x) - (1/2)csc^2 (x)
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