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

Paul Mitchell

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
$\underset{h\to 0}{lim}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
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,
$\frac{d}{dx}{f}^{-1}\left(x\right)={\left(\frac{df\left(x\right)}{dx}\right)}^{-1}$

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