## Sum of two metrics is a metric

It is quite straightforward to prove or verify that the sum of two metrics (distance functions) is still a metric.

Suppose $d_1(x,y)$ and $d_2(x,y)$ are metrics. Define $d(x,y)=d_1(x,y)+d_2(x,y)$.

## Positive-definite $d(x,y)\geq 0+0=0$. $d(x,y)=0 \iff d_1(x,y)=d_2(x,y)=0 \iff x=y$.

## Symmetry $d(x,y)=d_1(y,x)+d_2(y,x)=d(y,x)$.

## Triangle Inequality \begin{aligned} d(x,y) &\leq d_1(x,z)+d_1(z,y)+d_2(x,z)+d_2(z,y)\\ &=[d_1(x,z)+d_2(x,z)]+[d_1(z,y)+d_2(z,y)]\\ &=d(x,z)+d(z,y) \end{aligned}

Note that it follows by induction that the sum of any finite number of metrics (e.g. three, four or five metrics) is still a metric.

## Second-derivative Test For Extrema Of Functions Of Two Variables

Excerpt:

Proof of the second-derivative test. Our goal is to derive the second-derivative test, which determines the nature of a critical point of a function of two variables, that is, whether a critical point is a local minimum, a local maximum, or a saddle point, or none of these. In general for a function of n variables, it is determined by the algebraic sign of a certain quadratic form, which in turn is determined by eigenvalues of the Hessian matrix [Apo, Section 9.11]. This approach however relies on results on eigenvalues, and it may take several lectures to fully develop. Here we focus on the simpler setting when n = 2 and derive a test using the algebraic sign of the second derivative of the function.

The full proof can be found in the featured book below: T. Apostol, Calculus, vol. II, Second edition, Wiley, 1967

Featured book:  Calculus, Vol. 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability (Volume 2) # Proving Quotient Rule using Product Rule

This is how we can prove Quotient Rule using the Product Rule.

First, we need the Product Rule for differentiation: $\displaystyle\boxed{\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}}$

Now, we can write $\displaystyle\frac{d}{dx}(\frac{u}{v})=\frac{d}{dx}(uv^{-1})$

Using Product Rule, $\displaystyle \frac{d}{dx}(uv^{-1})=u(-v^{-2}\cdot\frac{dv}{dx})+v^{-1}\cdot(\frac{du}{dx})$

Simplifying the above will give the Quotient Rule! : $\displaystyle\boxed{\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}}$

You can also try proving Product Rule using Quotient Rule!