Hello its a me again! Today we will continue with** Mathematical Analysis** getting into **Derivatives**, **differentiation rules** and **common derivatives**. I leave out theorems and more in-depth examples for the next post! To understand better I suggest you to check out the Limits and Continuity post first. So, without further do let's get started!

# Derivatives:

Using the Limits and Continuity that we already talked about we can now get into the concept of derivatives. Derivatives are part of the **differential calculus** and are a **fundamental part of mathematical analysis **in general.

### Introduction:

So, why are they useful? The answer is simple, suppose you want to calculate the velocity of an object knowing the path/function that this object follows. The differentiation of his position with time will measure quick (**amount of change**) the position changes by time. The value of the **derivative at some point** actually tells us the** slope of the tangent line at that point**.

We actually already know the derivative of any point that is a part of a **linear function** by using the tangent of the angle ω between the line ε the linear function "draws" and the axis xx'.

So, the derivative is a = tan(ω) = y2 - y1 / x2 - x1.

This makes us think about the derivative as the **quotient of the change of y** (Δy) **and x (**Δx).

### Definition:

Well, does this mean that we can also apply the same for any function by saying:

*Derivative at (x0, f(x0)) = f(x) - f(x0) / x - x0 (where (x, f(x)) is some point "near" x0)* ??

No, it doesn't! Think about cases where we have a indeterminate form like 0/0 or ∞/∞.

So, tell me how did we solve indeterminate forms? Limits, you're right and this changes the definition having the quotient inside of a limit where x goes to x0.

Finally, we ended up with how we define a derivative.

The **definition **looks like this:

**f'(x0) = dy/dx = lim x->x0 [f(x) - f(x0) / x - x0]**

where f'(x0) is **Lagrange's notation** and dy/dx **Leibniz's notation**.

I prefer using Lagrange's in most cases!

### Differentiability at point:

Suppose a function f that is defined in the range (a, b) and a point x0 in that range.

This function is **differentiable at x0** only when the following **limit exists and has a real value**:

The** limit value** is called the **derivative of f at the point x0** and is noted as **f'(x0)**.

### Tangent at point:

Suppose a function f, the point A(x0, f(x0)) and that the derivative f'(x0) exists.

The **tangent of f at the point A** is a **line that contains A and has a slope f'(x0)**.

This means that the **tangent line equation** looks like this:

**y - f(x0) = f'(x0)*(x - x0)**

**Example:**

Suppose the function f(x) = -x - 4, when x<0

= x^2 - 4, when x>=0

- Is the function differentiable at the point x = 0?
- Can we define a tangent for the point A(2, f(2))?

We have that f(0) = 0^2 - 4 = -4.

The one-sided limits at x0 = 0 look like this:

These two limits are different and so the limit is non-existent and thus we have that the function f is non-differentiable at the point x = 0.

** Attention**! Don't confuse these limits with the continuity limits. We don't have to get a value equal to f(x0), but the two one-sided limits need to have the same value! (We will later on see that a function that is differentiable is also continuous at that point, but not all continuous functions are differentiable)

For the tangent line we simply have to calculate the derivative at point A and fill in the equation we talked about previous. So, f'(2) = 4 (we will talk about rules in a sec) and that way we have:

y - f(2) = 4(x - 2) => y - 0 = 4*x - 8 => y = 4*x - 8.

## Differentiation Rules:

- Multiplication with constant c -> [cf(x)]' = cf'(x)
- Sum/Difference of functions f, g -> [af(x) +- bg(x)]' = af'(x) +- bg'(x)
- Product of f, g -> [f(x)*g(x)]' = f'(x)g(x) + f(x)g'(x) (can be generalized to more, by having one of the functions in the product be "differentiated" at the time, followin the pattern)
- Quotient of f, g -> [f(x)/g(x)]' = (f'(x)g(x) − g'(x)f(x) ) / g(x)^2 (looks similar to the product one, but we this time have a minus and also divide with the denominator squared)
- A special case of the previous one is the so called Reciprocal -> [1/f(x)]' = −f'(x)/ f(x)^2
- Composition of f and g (chain rule) -> (fog(x))' = f'(g(x))*g'(x) that can also be written with Leibniz notation as dy/dx = (dy/du)*(du/dx), with y = f(u) and u = g(x), that makes us understand it better!

## Common Function Derivatives:

- Constant -> (c)' = 0
- Line -> (x)' = 1 or (ax)' = a
- Power -> (x^n)' = x^n*(x^n-1), so the square (x^2)' = 2x
- Square root -> [root(x)]' = [x^1/2]' = (1/2) * x^(-1/2) from the power rule
- Exponential -> (a^x)' = a^x * lna and so (e^x)' = e^x, cause lne = 1
- Logarithms -> [loga(x)]' = 1/(x*lna) and so lnx = 1/x, cause lne = 1
- Trigonometric and inverse functions that you can check out here
- Hyperbolic and inverse functions that you can check out here

## Examples:

We will calculate the derivative of the following functions:

**1. ** f(x) = x*e^x

We use the product rule and this gives us:

f'(x) = [x*e^x]' = x'e^x + x*(e^x)'.

The derivative of e^x is equal to itself and x' equals 1 and so we end up with:

f'(x) = e^x + x*e^x = e^x*(1 + x)

**2.** g(x) = x^2 / lnx

We use the quotinet rule and this gives us:

g'(x) = [x^2 / lnx]' = [(x^2)'*lnx - x^2*(lnx)'] / (lnx)^2

We know that the square derivative is equal to 2x and that lnx = 1/x and so:

g'(x) = [2x*lnx - x^2*(1/x)] / (lnx)^2

Simplify a little more and we get the final result:

g'(x) = (2x*lnx - x) / (lnx)^2

**3.** h(x) = arctan(5x^2) , where arctan is the inverse function of tan

We have a composition and so we will differentiate each part separately:

[arctan(5x^2)]'y = dy/du = 1/(1+y^2) = 1/(1+(5x^2)^2) = 1/(1+25x^4).

(5x^2)'x = du/dx = 2*5x = 10x.

So, our final result is:

[arctan(5x^2)]' = [1/(1+25x^4)] * 10x = 10x / (1+25x^4)

And this is actually it for today and I hope you enjoyed it!

Next time we will get into Differentiation Theorems and Examples as I already told you in the beginning of this post.

Until next time...Bye!