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?

Question

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?

Paul Mitchell · Accepted Answer

It’s all a matter of dependent and independent variables, when you wrote dθdtan⁡θ the denominator, that tan⁡θ, 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 tan⁡θ without setting tan⁡θ=x or tan⁡θ=a or to any other thing, the key point lies in the limit definition of the derivatives. Your expression dθdtan⁡θ cannot be put in the form  limh→0f(x+h)−f(x)h  until and unless we do some substitution for tan⁡θ   However, your substitution is correct, and we can find the same thing by the law of derivative of inverse function,  ddxf−1(x)=(df(x)dx)−1

levurdondishav4 · Accepted Answer

Heres

Showing \frac{d \theta}{d \tan \theta}=\frac{1}{1+\tan^2\theta} I suppose that \frac{d\theta}{d \tan \theta}=\frac{d \arctan

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