I want to find the total differential of $w$ from a given function
$w=\sqrt{u^2+v^2}\:with\:u\:=\:cos\left(ln\left(tan\left(p+\frac{1}{2}\pi \right)\right)\right)\:and\:v\:=\:sin\left(ln\left(tan\left(p+\frac{1}{2}\pi \:\right)\right)\right)$
to solving this problem I'm using
$\left(a\right)\:\:\:\:\frac{dw}{dp}=\frac{∂w}{∂u}\cdot \frac{du}{dp}+\frac{∂w}{∂v}\cdot \frac{dv}{dp}$
first I'm solving for the partial derivatives of $w$ with respect to $u$ and $v$ on RHS
$\left(b\right)\:\:\:\:\frac{∂w}{∂u}=\frac{u}{\sqrt{u^2+v^2}}\:\:\:\:and\:\:\:\:\frac{∂w}{∂v}=\frac{v}{\sqrt{u^2+v^2}}$
and then I'm solving for the total derivatives of $u$ and $v$ with respect to $p$ on RHS using chain rule.
$\frac{du}{dp}=\frac{d}{dx}cos\left(ln\left(tan\left(p+\frac{1}{2}\pi \right)\right)\right)=\frac{du}{dx}\cdot \frac{dx}{dy}\cdot \frac{dy}{dp}$
let $x=ln\left(tan\left(p+\frac{1}{2}\pi \:\right)\right)\:$ and $y\:=tan\left(p+\frac{1}{2}\pi \right)$
$\left(1\right)\:\frac{du}{dx}=\frac{dcos\left(x\right)}{dx}=-sin\left(x\right)=-sin\left(ln\left(tan\left(p+\frac{1}{2}\pi \right)\right)\right)$
$\left(2\right)\:\frac{dx}{dy}=\frac{dln\left(y\right)}{dy}=\frac{1}{y}=\frac{1}{tan\left(p+\frac{1}{2}\pi \right)}$
$\left(3\right)\:\frac{dy}{dp}=\frac{dtan\left(p+\frac{1}{2}\pi \right)}{dp}=sec^2\left(p+\frac{1}{2}\pi \right)$
substitute back $(1)$, $(2)$, and $(3)$ into $\frac{du}{dp}$ i get
$\left(c\right)\:\:\:\:\frac{du}{dp}=-\frac{sin\left(ln\left(tan\left(p+\frac{1}{2}\pi \right)\right)\right)}{tan\left(p+\frac{1}{2}\pi \:\right)cos^2\left(p+\frac{1}{2}\pi \:\right)}$
now i want to find $\frac{dv}{dp}$
$\frac{dv}{dp}=\frac{dsin\left(ln\left(tan\left(p+\frac{1}{2}\pi \right)\right)\right)}{dp}=\frac{dv}{dx}\cdot \frac{dx}{dy}\cdot \frac{dy}{dp}$
since $\frac{dx}{dy}$ and $\frac{dy}{dp}$ have the same result as $(2)$ and $(3)$, now i only have to find $\frac{dv}{dx}$ with $v\:=\:sin\left(x\right)$ i get $\frac{dv}{dx}=cos\left(x\right)=cos\left(ln\left(tan\left(p+\frac{1}{2}\pi \right)\right)\right)$ and then substitute them back into $\frac{dv}{dp}$ yields
$\left(d\right)\:\:\:\:\frac{dv}{dp}=\frac{cos\left(ln\left(tan\left(p+\frac{1}{2}\pi \right)\right)\right)}{tan\left(p+\frac{1}{2}\pi \:\right)cos^2\left(p+\frac{1}{2}\pi \:\right)}$
finally, substitute $(b)$, $(c)$, and $(d)$ into $(a)$ i get
$\frac{dw}{dp}=\frac{cot\left(p+\frac{1}{2}\pi \:\right)\cdot sec^2\left(p+\frac{1}{2}\pi \:\right)}{\sqrt{u^2+v^2}}\left(v\cdot cos\left(ln\left(tan\left(p+\frac{1}{2}\pi \:\right)\right)\right)-u\cdot sin\left(ln\left(tan\left(p+\frac{1}{2}\pi \:\:\right)\right)\right)\right)$
this is my answer, but i'm not sure is there an error in my calculation? and how to know if my answer is correct? thank you for your time and advice, it has nothing to do with homework or school, i just have no one to ask to correct my answer.
If we denote $$\phi := \ln \tan \left(p + \frac{\pi}{2}\right),$$ then $$w = \sqrt{\cos^2 \phi + \sin^2 \phi} = 1,$$ and thus $$\frac{dw}{dp} = 0.$$