David H. answered • 02/06/17
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1.) what is the derivative of cot(x)?
multiple ways to get but one way is to use quotient rule on cot(x) = cosx/sinx.
you should get -csc^2(x).
2.) you can then use chain rule .
hope helpful. Let me know if u have issues with chain rule.
best -david
David H.
I don't think so.
Derivative of 15cot(-1t) = 15(-csc2(-1t))*(-1) = 15csc2(-1t)
derivative of 15cot(-1/t) = 15(-csc2(-1/t))*(1/t2) = -15csc2(-1/t)*(1/t2)
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02/06/17
Michael J.
This is an inverse cotangent, rather than tangent. The derivative of the inverse is
dy/dx[cot-1(t)] = 1 / (1 + t2)
Note: We are not taking the derivative of cotangent, but its inverse just as I shown above. Also, the final answer should not be a numerical value, because your not evaluating the derivative at a certain point. It is only a general derivative.
Ketevan, I gave you a solution to this problem the other day. Did you read it?
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02/06/17
Michael J.
Actually, the derivative is
dy/dx[cot-1(t)] = - 1 / (1 + t2)
Had to make that correction. There should be a negative sign in the derivative.
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02/06/17
David H.
Yes ketevan if this is meant to be the inverse cotangent then use the derivative of that plus chain rule and you should have your answer.
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02/06/17
Ketevan M.
02/06/17