Final answer:
The derivative of cot^-1u is -1/(u^2+1) times the derivative of u with respect to x, computed using the chain rule and inverse trigonometric derivative properties.
Step-by-step explanation:
The derivative of the inverse cotangent function, denoted as cot-1u, is found using the chain rule and properties of derivatives of inverse trigonometric functions. To differentiate cot-1u, let y = cot-1u. Then, taking the cotangent of both sides gives cot(y) = u. Differentiating this relation implicitly with respect to x gives -csc2y (dy/dx) = du/dx.
This implies dy/dx = -1/(u2+1) (du/dx), given that csc2(y) = 1 + cot2(y), which translates to 1 + u2 after substitution.
Therefore, the derivative of cot-1u with respect to x is -1/(u2+1) times the derivative of u with respect to x.
The derivative of the function cot-1u can be found using the chain rule. Let's denote the cot-1 function as y and the variable u as x. So, we have y = cot-1(x). To find the derivative, we need to first find dy/dx.
We know that cot-1(x) is the inverse of cot(x). The derivative of cot(x) is -csc2(x). Therefore, the derivative of cot-1(x) is -1/(-csc2(x)), which simplifies to 1/(csc2(x)).
Substituting u back into the equation, the derivative of cot-1u is 1/(csc2u).