Ketevan M.

# calculus 2

h(t) = 15 cot−1(t) + 15 cot−1(1/t)

finding derivative

By:

Tutor
5 (18)

Patient & Effective Math/Physics Tutor near Fort Worth

Ketevan M.

As final answer I got 0. Is 0 right?
Report

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)
Report

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?
Report

02/06/17

Michael J.

Actually, the derivative is

dy/dx[cot-1(t)] = - 1 / (1 + t2)

Report

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.
Report

02/06/17

## Still looking for help? Get the right answer, fast.

Get a free answer to a quick problem.
Most questions answered within 4 hours.

#### OR

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.