The derivative of cotangent, d/dx(cotx), is proven to be -csc^2x using the chain rule. The discussion outlines the steps taken, starting with the function f(x) = cotx expressed as (tanx)^(-1). The proof involves calculating f'(x) and g'(x) for f(x) and g(x) = tanx, leading to the final result of -csc^2x. The solution is confirmed as valid by participants in the discussion.
PREREQUISITESStudents studying calculus, particularly those focusing on derivatives of trigonometric functions, as well as educators looking for clear examples of applying the chain rule.
It's not a good idea to use "x" as the variable here. Use, say, u instead:dylanhouse said:Homework Statement
Using the chain rule, prove that d/dx(cotx)= -csc^2x
Homework Equations
Chain rule
The Attempt at a Solution
Is this correct?
f(x)=cotx=(tanx)^(-1)
Let f(x) = (x)^-1 Therefore, f'(x)= -1/(x^2)
Yes, that is a valid proof.Let g(x) = tanx Therefore, g'(x)= sec^2x
F'(x)=f'(g(x))g'(x)
=(-1/(tanx)^2)(secx^2)
=(-secx^2)/(tanx^2)
= -csc^2 x