Showing dθdtan⁡θ=11+tan2θ I suppose that dθdtan⁡θ=darctan⁡xdx=11+x2=11+tan2θ So is dθdtan⁡θ=11+tan2θ correct? And dθdtan⁡θ2=21+tan2θ2?

William Boggs

William Boggs

Answered question


Showing dθdtanθ=11+tan2θ
I suppose that
So is
correct? And

Answer & Explanation

Paul Mitchell

Paul Mitchell

Beginner2022-01-17Added 40 answers

It’s all a matter of dependent and independent variables, when you wrote dθdtanθ the denominator, that tanθ, became a independent variable and θ became dependent variable.
But it is quite unconventional to do that, because you see you cannot obatain θ from tanθ without setting tanθ=x or tanθ=a or to any other thing, the key point lies in the limit definition of the derivatives. Your expression dθdtanθ cannot be put in the form
until and unless we do some substitution for tanθ
However, your substitution is correct, and we can find the same thing by the law of derivative of inverse function,


Beginner2022-01-18Added 38 answers


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