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Ketevan M.

asked • 02/06/17

calculus 2

h(t) = 15 cot−1(t) + 15 cot−1(1/t)
 
finding derivative

1 Expert Answer

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David H. answered • 02/06/17

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Ketevan M.

As final answer I got 0. Is 0 right?
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02/06/17

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

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