William Boggs

2022-01-16

Showing $\frac{d\theta}{d\mathrm{tan}\theta}=\frac{1}{1+{\mathrm{tan}}^{2}\theta}$

I suppose that

$\frac{d\theta}{d\mathrm{tan}\theta}=\frac{d\mathrm{arctan}x}{dx}=\frac{1}{1+{x}^{2}}=\frac{1}{1+{\mathrm{tan}}^{2}\theta}$

So is

$\frac{d\theta}{d\mathrm{tan}\theta}=\frac{1}{1+{\mathrm{tan}}^{2}\theta}$

correct? And

$\frac{d\theta}{d\mathrm{tan}\frac{\theta}{2}}=\frac{2}{1+{\mathrm{tan}}^{2}\frac{\theta}{2}}$ ?

I suppose that

So is

correct? And

Paul Mitchell

Beginner2022-01-17Added 40 answers

It’s all a matter of dependent and independent variables, when you wrote $\frac{d\theta}{d\mathrm{tan}\theta}$ the denominator, that $\mathrm{tan}\theta$ , became a independent variable and $\theta$ became dependent variable.

But it is quite unconventional to do that, because you see you cannot obatain$\theta$ from $\mathrm{tan}\theta$ without setting $\mathrm{tan}\theta =x$ or $\mathrm{tan}\theta =a$ or to any other thing, the key point lies in the limit definition of the derivatives. Your expression $\frac{d\theta}{d\mathrm{tan}\theta}$ cannot be put in the form

$\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

until and unless we do some substitution for$\mathrm{tan}\theta$

However, your substitution is correct, and we can find the same thing by the law of derivative of inverse function,

$\frac{d}{dx}{f}^{-1}\left(x\right)={\left(\frac{df\left(x\right)}{dx}\right)}^{-1}$

But it is quite unconventional to do that, because you see you cannot obatain

until and unless we do some substitution for

However, your substitution is correct, and we can find the same thing by the law of derivative of inverse function,

levurdondishav4

Beginner2022-01-18Added 38 answers

Heres