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

Source: http://www.math.illinois.edu/~verahur/18.024/notesSD.pdf

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

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!