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(-csc

^{2}(-1t))*(-1) = 15csc^{2}(-1t)derivative of 15cot(-1/t) = 15(-csc

^{2}(-1/t))*(1/t^{2}) = -15csc^{2}(-1/t)*(1/t^{2})
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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 + t^{2})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 + t^{2})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