]> Definition and Rules

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Chapter 1: Differentiation; Section 3: Derivatives; page 1

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Definition and Rules

Definition of a derivative

Differentiation and integration, which deal with derivatives and integrals accordingly, are called calculus and were invented in the 17^th century by I. Newton, in order to develop his theory of physics. Since then calculus has become an important subject of mathematics and serves as a milestone for many more further developments.

The derivative of a single valued continuous function $y = f (x)$ at a given point $x_{0}$ is the limit

$\frac{d f}{d x} = \lim_{Δ x \to 0} \frac{Δ f}{Δ x} = \lim_{x \to x_{0}} \frac{f (x) - f (x_{0})}{x - x_{0}}$	(1.3.1.1)

The two expressions with the limits are obviously identical, just presented differently. The left-hand side expression $(\frac{d f}{d x})$ represents the limit and not a ratio of two numbers. It signifies that it originates from an increment of f divided by an increment of x . In case of physical dimensions, it also stresses the ratio of the corresponding dimensions.

The derivative is usually represented as f ' or y ' [for y = f(x)] , but we are not going to use this notation in order to avoid ambiguity.

If a function is not single valued, one has to specify a single valued branch. The continuity of the function is a necessary condition, but not a sufficient one, e.g. if at (1.3.1.1) the limit from the right is not identical to the limit from the left, the function is not differentiable at this point. The case of the limit diverging to infinty also makes the function not differentiable at that point.

We already know (1.2.5.14) that for a linear function (corresponding to a straight line) the ratios of the increments (1.3.1.1) are constant and correspond to the slope. The meaning of the limit (1.3.1.1) is the slope at a particular point (x₀).

In order to understand the notion and the meaning of a derivative, we'll start by looking at one particular example: the function.

$y = \sqrt{x} at x_{0} = 1$	(1.3.1.2)

The function is single valued and continuous. The calculation of the derivative in this case is:

${(\frac{d y}{d x})}_{x_{0}} = \lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1} = \lim_{x \to 1} \frac{\sqrt{x} - 1}{(\sqrt{x} - 1) (\sqrt{x} + 1)} = \lim_{x \to 1} \frac{1}{\sqrt{x} + 1} = \frac{1}{2}$	(1.3.1.3)

If we know the slope of the function at a point, then we can define the tangent line to the function as a straight line passing through that point with this slope. According to (1.2.5.18) it is

$y = y_{0} + {(\frac{d y}{d x})}_{0} (x - x_{0})$	(1.3.1.4)

where ${(\frac{d y}{d x})}_{0}$ means the derivative at the point $(x_{0}, y_{0})$ of the function.

In our case we obtain $y = 0.5 (x + 1)$ . This particular example is explained graphically at Fig. Derivative by limit.

From the definition of the tangent line at a point, in the vicinity of the point the function and the tangent line almost overlap and the relation (1.3.1.4) is also called the first approximation of the function around this point. In other words if we could zoom-in enough, the graphic presentation of a function around this point, we'll see it approximated by a straight line, representing the tangent line. The same example is used again for showing this feature at Fig. Derivative by zoom.

We already know from the discussion about straight lines (previous section), that in a graphic presentation with equally scaled coordinate axes, the slope is by value equivalent to the tangent of the angle made with the x axis. This was the case in the Fig-window just shown, but one has to be careful and use the tangent angle interpretation of the slope with caution.

It is obvious that the slope of a linear function (straight line) is its derivative, and it is constant. We'll use this to show a simple example of a function that is continuous in the whole domain of numbers, but is not differentiable at a point. The function

$y = \| x \| = {\begin{cases} - x for x \leq 0 \\ x for x \geq 0 \end{cases}}$

is not differentiable at point x = 0 since the derivative from the left is −1 and - from the right is +1 . This means that a tangent line at that point is not defined, and that the function has a kink there.

Another way to denote a derivative is

$\frac{d}{d x} f (x) = \lim_{Δ x \to 0} \frac{f (x + Δ x) - f (x)}{Δ x}$	(1.3.1.5)

where the meaning of the ratio $\frac{d}{d x}$ is an operator applied to the function. This notation is very useful when the function is differentiable in a whole domain, and the derivative itself is a function. We'll deal in particular with such functions.

A direct consequence of the definition (1.3.1.5) is that the derivative of a constant is zero.

$\frac{d}{d x} C = 0 where C = constant$	(1.3.1.6)

Rules for differentiation and $\frac{d}{d x} (x^{p})$

We'll start with the differentiation of $x^{p}$ , where p is a natural number. In the case of p = 2

$\frac{d}{d x} x^{2} = \lim_{Δ x \to 0} \frac{{(x + Δ x)}^{2} - x^{2}}{Δ x} = \lim_{Δ x \to 0} \frac{2 x Δ x + {(Δ x)}^{2}}{Δ x} = 2 x$	(1.3.1.7)

where the properties of the binomial expansion (1.1.2.20) is used. In the general case one has for any natural p

${(x + Δ x)}^{p} = (x^{p} + p x^{p - 1} Δ x + ....)$

where the missing terms cannot contribute to the derivative as seen from (1.3.1.7). One obtains therefore for p = natural

$\frac{d}{d x} x^{p} = p x^{p - 1}$	(1.3.1.8)

This includes also the case of p = 0 as seen from (1.3.1.6).

Rule 1 for differentiation

$\frac{d}{d x} (\frac{1}{f (x)}) = - \frac{\frac{d f}{d x}}{f^{2}}$	(1.3.1.9)

Proof:

$\frac{d}{d x} (\frac{1}{f (x)}) = \lim_{Δ x \to 0} \frac{\frac{1}{f (x + Δ x)} - \frac{1}{f (x)}}{Δ x} = \lim_{Δ x \to 0} - \frac{f (x + Δ x) - f (x)}{Δ x f (x) f (x + Δ x)}$

and one obtains (1.3.1.9).

One can extend (1.3.1.8) to any integer p . Let's take p = −n , where n is a natural number, than according to (1.3.1.9) and (1.3.1.8):

$\frac{d}{d x} (x^{p}) = \frac{d}{d x} (\frac{1}{x^{n}}) = - \frac{n x^{n - 1}}{x^{2 n}} = p x^{- n - 1} = p x^{p - 1}$

that extends (1.3.1.8) to p = integer.

Rule 2 for differentiation

$\frac{d y}{d x} = \frac{1}{\frac{d x}{d y}} where {\begin{cases} y = inv f (x) \\ x = f (y) \end{cases}}$	(1.3.1.10)

Proof: if y and x are defined as in (1.3.1.10), then there is one to one correspondance of any point (x,y) of the function to a point of the inverse-function. Therefore, any choice of Δx and Δy of the function matches an equal pair of the inverse-function. We obtain

$\begin{array}{l} \frac{d}{d x} inv f (x) = \frac{d y}{d x} = \lim_{Δ x \to 0} \frac{inv f (x + Δ x) - inv f (x)}{Δ x} = \lim_{Δ x \to 0} \frac{Δ y}{Δ x} = \\ = \lim_{Δ y \to 0} \frac{1}{\frac{Δ x}{Δ y}} = \lim_{Δ y \to 0} \frac{1}{\frac{f (y + Δ y) - f (y)}{Δ y}} = \frac{1}{\frac{d x}{d y}} = \frac{1}{\frac{d}{d y} f (y)} \end{array}}$

With the aid of (1.3.1.10), one can extend (1.3.1.8) to also include p = 1/n , where n is an integer. The function and its inverse are

$x = y^{n} and y = x^{\frac{1}{n}} where p = \frac{1}{n}$

and from (1.3.1.10) we have

$\frac{d}{d x} x^{p} = \frac{1}{\frac{d}{d y} y^{n}} = \frac{1}{n y^{n - 1}} = \frac{y}{n x} = p x^{p - 1}$

In case of double valued root (y positive or negative), we did not specify the sign of y and the equation is therefore correct in either way.

Rule 3 for differentiation

$\frac{d y}{d x} = \frac{d f}{d u} \frac{d u}{d x} where {\begin{cases} y = f (u) \\ u = u (x) \end{cases}}$	(1.3.1.11)

is commonly called the chain rule.

Proof: For a given Δx one obtains a corresponding Δu , for it one obtains a corresponding Δy = Δf .

$\frac{d y}{d x} = \lim_{Δ x \to 0} \frac{Δ y}{Δ x} = \lim_{Δ x \to 0} \frac{Δ y Δ u}{Δ x Δ u} = \lim_{Δ u \to 0} \frac{Δ y}{Δ u} \lim_{Δ x \to 0} \frac{Δ u}{Δ x} = \frac{d f}{d u} \frac{d u}{d x}$

By the use of the chain rule (1.3.1.11) one can extend (1.3.1.8) to p =rational = m/n , where m and n integers (n ≠ 0). From

$or {\begin{cases} y = x^{p} = x^{\frac{m}{n}} \\ y = f (u) = u^{m} and u = x^{\frac{1}{n}} \end{cases}$

and with the use of (1.3.1.11) one obtains

$\begin{array}{l} \frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x} = m u^{m - 1} \frac{1}{n} x^{(\frac{1}{n} - 1)} = \\ = m x^{\frac{m - 1}{n}} \frac{1}{n} x^{(\frac{1}{n} - 1)} = \frac{m}{n} x^{(\frac{m}{n} - \frac{1}{n} + \frac{1}{n} - 1)} = p x^{p - 1} \end{array}}$

The extension of (1.3.1.8) for p =real is obtained from the continuity.

Rule 4 for differentiation

$\frac{d y}{d x} = C \frac{d}{d x} f (x) where {\begin{cases} y = C f (x) \\ C = const \end{cases}}$	(1.3.1.12)

Proof:

$\begin{array}{l} \frac{d y}{d x} = \lim_{Δ x \to 0} \frac{y (x + Δ x) - y (x)}{Δ x} = \lim_{Δ x \to 0} \frac{C f (x + Δ x) - C f (x)}{Δ x} = \\ = C \lim_{Δ x \to 0} \frac{f (x + Δ x) - f (x)}{Δ x} = C \frac{d f}{d x} \end{array}}$

Scaling a function by a factor results in scaling the derivative by the same factor. This allows one to use the same graphic display for different scales of a function just by making the appropriate change in the scale of the ordinate. The following graphics clarifies that point Fig. Scale a function

An example of this rule (1.3.1.12):

$\begin{array}{l} for y = x^{3} \frac{d y}{d x} = 3 x^{2} \\ for y = 5 x^{3} \frac{d y}{d x} = 15 x^{2} \end{array}}$

Rule 5 for differentiation

$\frac{d}{d x} (f_{1} (x) + f_{2} (x)) = \frac{d}{d x} f_{1} (x) + \frac{d}{d x} f_{2} (x)$	(1.3.1.13)

Proof:

$\begin{array}{l} \frac{d}{d x} (f_{1} (x) + f_{2} (x)) = \lim_{Δ x \to 0} \frac{f_{1} (x + Δ x) + f_{2} (x + Δ x) - f_{1} (x) - f_{2} (x)}{Δ x} = \\ = \lim_{Δ x \to 0} \frac{f_{1} (x + Δ x) - f_{1} (x)}{Δ x} + \lim_{Δ x \to 0} \frac{f_{2} (x + Δ x) - f_{2} (x)}{Δ x} = \frac{d}{d x} f_{1} (x) + \frac{d}{d x} f_{2} (x) \end{array}}$

At this stage we can differentiate a polynomial:

$\frac{d}{d x} (\frac{x^{3}}{9} - x) = \frac{3 x^{2}}{9} - 1 = \frac{x^{2}}{3} - 1$	(1.3.1.14)

or using the general definition of a polynomial of order n (1.2.2.6):

$\frac{d}{d x} p_{n} (x) = \frac{d}{d x} \sum_{k = 0}^{n} a_{k} x^{n - k} = \sum_{k = 0}^{n - 1} a_{k} (n - k) x^{n - 1 - k}$	(1.3.1.15)

The last term (k = n) of the final result at (1.3.1.15) was omited, since it corresponds to the derivative of a constant, which is zero. It is obvious that the differentiation of a polynomial reduces its order by one unit.

Rule 6 for differentiation

$\frac{d}{d x} (f_{1} f_{2}) = \frac{d f_{1}}{d x} f_{2} + \frac{d f_{2}}{d x} f_{1} where {\begin{cases} f_{1} = f_{1} (x) \\ f_{2} = f_{2} (x) \end{cases}}$	(1.3.1.16)

Proof:

$\begin{array}{l} \frac{d}{d x} (f_{1} (x) f_{2} (x)) = \lim_{Δ x \to 0} \frac{f_{1} (x + Δ x) f_{2} (x + Δ x) - f_{1} (x) f_{2} (x)}{Δ x} = \\ = \lim_{Δ x \to 0} \frac{f_{1} (x + Δ x) f_{2} (x + Δ x) - f_{1} (x) f_{2} (x + Δ x) + f_{1} (x) f_{2} (x + Δ x) - f_{1} (x) f_{2} (x)}{Δ x} = \\ = \lim_{Δ x \to 0} \frac{f_{1} (x + Δ x) f_{2} (x + Δ x) - f_{1} (x) f_{2} (x + Δ x)}{Δ x} + \lim_{Δ x \to 0} \frac{f_{1} (x) f_{2} (x + Δ x) - f_{1} (x) f_{2} (x)}{Δ x} = \\ = \lim_{Δ x \to 0} \frac{f_{1} (x + Δ x) - f_{1} (x)}{Δ x} f_{2} (x + Δ x) + \lim_{Δ x \to 0} \frac{f_{2} (x + Δ x) - f_{2} (x)}{Δ x} f_{1} (x) = \\ = \frac{d f_{1}}{d x} f_{2} (x) + \frac{d f_{2}}{d x} f_{1} (x) \end{array}}$

\begin{array}{l} \frac{d}{d x} (f_{1} (x) f_{2} (x)) = \lim_{Δ x \to 0} \frac{f_{1} (x + Δ x) f_{2} (x + Δ x) - f_{1} (x) f_{2} (x)}{Δ x} = \\ = \lim_{Δ x \to 0} \frac{f_{1} (x + Δ x) f_{2} (x + Δ x) - f_{1} (x) f_{2} (x + Δ x) + f_{1} (x) f_{2} (x + Δ x) - f_{1} (x) f_{2} (x)}{Δ x} = \\ = \lim_{Δ x \to 0} \frac{f_{1} (x + Δ x) f_{2} (x + Δ x) - f_{1} (x) f_{2} (x + Δ x)}{Δ x} + \lim_{Δ x \to 0} \frac{f_{1} (x) f_{2} (x + Δ x) - f_{1} (x) f_{2} (x)}{Δ x} = \\ = \lim_{Δ x \to 0} \frac{f_{1} (x + Δ x) - f_{1} (x)}{Δ x} f_{2} (x + Δ x) + \lim_{Δ x \to 0} \frac{f_{2} (x + Δ x) - f_{2} (x)}{Δ x} f_{1} (x) = \\ = \frac{d f_{1}}{d x} f_{2} (x) + \frac{d f_{2}}{d x} f_{1} (x) \end{array}}

As an example, the polynomial from (1.3.1.14) will be used, but expressed as a product of two functions:

$\frac{d}{d x} (\frac{x^{3}}{9} - x) = \frac{d}{d x} (\frac{x}{9} (x^{2} - 9)) = \frac{x^{2} - 9}{9} + \frac{x}{9} 2 x = \frac{x^{2}}{3} - 1$

Rule 7 for differentiation

$\frac{d}{d x} (\frac{f_{1}}{f_{2}}) = \frac{\frac{d f_{1}}{d x} f_{2} - f_{1} \frac{d f_{2}}{d x}}{f_{2}^{2}} where {\begin{cases} f_{1} = f_{1} (x) \\ f_{2} = f_{2} (x) \end{cases}}$	(1.3.1.17)

Proof: The proof is based on the rules 1 (1.3.1.9) and 3 (1.3.1.11).

$\frac{d}{d x} (\frac{f_{1}}{f_{2}}) = \frac{d}{d x} (f_{1} \frac{1}{f_{2}}) = \frac{d f_{1}}{d x} \frac{1}{f_{2}} - f_{1} \frac{\frac{d f_{2}}{d x}}{f_{2}^{2}} = \frac{\frac{d f_{1}}{d x} f_{2} - f_{1} \frac{d f_{2}}{d x}}{f_{2}^{2}}$

As an example the derivative of a ratio of two polynomials will be calculated:

$\frac{d}{d x} (\frac{x^{2} - 1}{x^{2} + 1}) = \frac{2 x (x^{2} + 1) - 2 x (x^{2} - 1)}{{(x^{2} + 1)}^{2}} = \frac{4 x}{{(x^{2} + 1)}^{2}}$

Parity of the derivative

From the definition of a derivative (1.3.1.5) one can prove that the differentiation inverts the parity of a function. It means that if the parity of a function is defined as even(odd), the parity of the derivative is odd(even). The proof is given as an exercise here below. (1.3.1.14) can be used as an example of the rule.

Exercises

Exercise 1. Prove that differentiation inverts the parity of a function. Hint: use (1.3.1.5) and see separately the cases of an even and an odd function.

______________

Find the derivatives of the following functions. If applicable check if the derivative inverts the parity of the function.

Exercise 2.

$y = \frac{1}{x - \sqrt{1 - x^{2}}}$

Exercise 3.

$y = {(x^{8} + 3 x^{4} - x^{2})}^{6}$

Exercise 4.

$y = \frac{\sqrt{x} + 1}{\sqrt{x} - 1}$

Exercise 5.

$y = {(\frac{x^{2} + a x + a^{2}}{x^{2} - a x + a^{2}})}^{3} where a = constant$

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Math Animated™ Mathematical Introduction for Physics and Engineering by Samuel Dagan (Copyright © 2007-2020)

Chapter 1: Differentiation; Section 3: Derivatives; page 1

Home | Table of Contents | A-Z Index | Help

Definition and Rules

Definition of a derivative

Rules for differentiation and d dx ( x p )

Parity of the derivative

Exercises

Home | Table of Contents | A-Z Index | Help

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Rules for differentiation and $\frac{d}{d x} (x^{p})$