What is the norm of this vector? I have the vector:
$$(-\sin(\varphi)\cos(\theta) \varphi' - \cos(\varphi)\sin(\theta)\theta', \cos(\varphi)\sin(\theta)\varphi' + \sin(\varphi)\cos(\theta)\theta',\cos(\theta)\theta') $$
where $\varphi=\varphi(t) ; \theta = \theta(t);$ and $\varphi'= \frac{d \varphi}{dt};\theta'= \frac{d \theta}{dt};$
I have the answer to be : $\sqrt{1+\varphi'^2\cos^2\theta}$.
I have tried and tried simplifying this to get the answer, but I cannot seem to. Does anyone see something that I cannot, how to get this?   
 A: It has to be the derivative of $\vec r=(\cos\varphi(t)\cos\theta(t),\sin\varphi(t)\cos\theta(t),\sin\theta(t))$ if you want to get the given solution.
$\dfrac{d\vec r}{dt}=\dfrac{\partial\vec r}{\partial\theta}\dfrac{d\theta}{dt}+\dfrac{\partial\vec r}{\partial\varphi}\dfrac{d\varphi}{dt}=$
$=(-\varphi'\sin\varphi\cos\theta-\theta'\cos\varphi\sin\theta,\varphi'\cos\varphi\cos\theta -\theta'\sin\varphi\sin\theta,\theta'\cos\theta)$
$\require{enclose}\enclose{downdiagonalstrike,updiagonalstrike}{\left|\dfrac{d\vec r}{dt}\right|=\sqrt{1+\varphi'^2\cos^2\theta}}$
added
With the detailed calculations, the given solution doesn't follow, at most $\sqrt{\theta'^2+\varphi'^2\cos\theta}$. We can recover the given solution redefining $\varphi'$ as $\dfrac{d\varphi}{d\theta}$, but to many assumptions. I don't delete the answer because maybe some calculations are usefull.
$$\left|\dfrac{d\vec r}{dt}\right|^2=x^2+y^2+z^2=$$
$$=(-\varphi'\sin\varphi\cos\theta-\theta'\cos\varphi\sin\theta)^2+(\varphi'\cos\varphi\cos\theta -\theta'\sin\varphi\sin\theta)^2+(\theta'\cos\theta)^2=$$ $$=\varphi'^2\sin^2\varphi\cos^2\theta+\theta'^2\cos^2\varphi\sin^2\theta+2\theta'\cos\varphi\sin\theta\;\varphi'\sin\varphi\cos\theta+$$
$$+\varphi'^2\cos^2\varphi\cos^2\theta+\theta'^2\sin^2\varphi\sin^2\theta-2\varphi'\cos\varphi\cos\theta\;\theta'\sin\varphi\sin\theta+\theta'^2\cos^2\theta=$$
$$=\varphi'^2\cos^2\theta(\sin^2\varphi+\cos^2\varphi)+\theta'^2\sin^2\theta(\sin^2\varphi+\cos^2\varphi)+\theta'^2\cos^2\theta=$$
$$=\varphi'^2\cos^2\theta+\theta'^2(\sin^2\theta+\cos^2\theta)=\theta'^2+\varphi'^2\cos\theta$$
