AP Calculus AB and BC

Derivative of arccot x: Answer and Sign

The derivative of arccot x is negative 1 over 1 plus x squared. It is the exact negative of the arctangent derivative, because arctan x plus arccot x equals pi over 2, and differentiating a constant gives zero.

ddx[arccot⁡x]=−11+x2\frac{d}{dx}\left[\operatorname{arccot} x\right] = -\frac{1}{1+x^{2}}

From the complementary identity

arctan⁡x+arccot⁡x=π2\arctan x + \operatorname{arccot} x = \frac{\pi}{2}

Differentiating both sides gives zero on the right, so the two derivatives must cancel. That is faster than implicit differentiation and explains the sign rather than asking you to memorise it.

ddxarccot⁡x=−11+x2\frac{d}{dx}\operatorname{arccot} x = -\frac{1}{1+x^{2}}

The co-function pattern again

Every inverse co-function derivative is the negative of its partner: arccos against arcsin, arccot against arctan, arccsc against arcsec. One rule covers all three pairs.

Common mistakes

  • Dropping the minus sign, which gives the arctangent derivative.
  • Confusing arccot⁡x\operatorname{arccot} x with cot⁡x\cot x or with 1arctan⁡x\frac{1}{\arctan x}.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

What is the derivative of arccot x?

It is −11+x2-\frac{1}{1+x^{2}}.

Why is it negative?

Because arctan⁡x+arccot⁡x\arctan x + \operatorname{arccot} x is the constant π2\frac{\pi}{2}, so the derivatives must sum to zero.