Hi! This is just a short introduction to how you would prove some of the various rules used in calculus to differentiate equations using first principles. The rules that will be discussed include:

  • Power rule
  • Product rule
  • Quotient rule

The following first principles example is for establishing the power rule.

Case 1

Begin with $y = x^2$; Fundamental notion of calculus is ‘growing’. Now, as y and $x^2$ are equal to one another, it is clear that if x grows, $x^2$ will also grow. And if $x^2$ grows, so will y.

What we’re looking for is the proportion between the growing of y and the growing of x i.e. the ratio of dy and dx, the value of $\frac{dy}{dx}$

Let x, then, grow a little bit bigger and become $x + dx$; similarly, y will grow a bit bigger and become $y + dy$ therefore:

$$y + dy = (x + dx)^2$$

Doing the squaring:

$$y + dy = x^2 + 2x \cdot dx + (dx)^2$$

What is $(dx)^2$? Remember that dx means a little bit of x. Then $(dx)^2$ will mean a little bit of a little bit (smaller in the second magnitude) and can therefore be disregarded. Leaving it out we then get:

$$y + dy = x^2 + 2x \cdot dx$$

Now $y = x^2$; So let us subtract this from the equation:

$$dy = 2x \cdot dx$$

Dividing across by dx, we find:

$$\frac{dy}{dx} = 2x$$

This is what we set out to find, the ratio of the growing of y to the growing of x.

Case 2

Differentiating $y = x^3$ using first principles.

$$y + dy = (x + dx)^3$$

Cubing out:

$$y + dy = x^3 + 3x^2 \cdot dx + 3x(dx)^2 + (dx)^3$$

Subtract the original ($y = x^3$) and remove the negligible values(smallness in the second magnitude):

$$dy = 3x^2 \cdot dx$$

And divide both sides by dx

$$\frac{dy}{dx} = 3x^2$$

What this shows, is that the differential of $y = x^2$ is equal to $\frac{dy}{dx} = 2x$, etc. This is the basis of the one of the rules of differentiation, the power rule


$y = x^2 \space$differentiated $\frac{dy}{dx} = 2x$

$y = x^3 \space$differentiated $\frac{dy}{dx} = 3x^2$

$y = x^4 \space$differentiated $\frac{dy}{dx} = 4x^3$


Power rule

If we want to deal with any higher power, say n - $y = x^n$; Then we should expect to find $\frac{dy}{dx} = nx^{\left( {n-1} \right)}$.

$$\frac{dy}{dx} = nx^{\left( {n-1} \right)} $$

Power rule exceptions, and the like

Dealing with negative and fractional equations:


$y = x^\frac{1}{2} \space$differentiated $\frac{dy}{dx} = \frac{1}{2}x^{{-\frac{1}{2}}}$


$y = x^{-2} \space$differentiated $\frac{dy}{dx} = -2x^{-3}$

Dealing with multiplied constants

let $y = 7x^2$

$$dy + y = 7(x + dx)^2$$

$$dy + y = 7[x^2 + 2x \cdot dx + (dx)^2]$$

$$dy + y = 7x^2 + 14x \cdot dx + 7(dx)^2$$

Subtract the original equation ( $y = 7x^2$ ) and disregard values of 2nd order magnitudes

$$dy = 14x \cdot dx$$

Then divide by dx

$$\frac{dy}{dx} = 14x$$

$\therefore \frac{dy}{dx} = a \cdot nx^{n-1}$

This is true for division as well.

Dealing with the sum of two or more functions; You’d simply differentiate them one after another.

$$let \space y = (x^2 + c) + (ax^4 + b)$$

$$then \space \frac{dy}{dx} = 2x + 4ax^3$$

A note on dealing with constants

$$y = x^3 + 5$$


$$y + dy = (x + dx)^3 + 5$$

$$y + dy = x^3 + 3x^2dx + 3x(dx)^2 + (dx)^3 + 5$$

Negating the small quantities of higher orders of magnitude, this becomes:

$$y + dy = x^3 + 3x^2 \cdot dx + 5$$

Subtract the original equation ($y = x^3 + 5$)

$$dy = 3x^2 \cdot dx$$


$$\frac{dy}{dx} = 3x^2$$

So we see - when subtracting the original equation, it removes the constants. When following the power rule, we can therefore omit the constant values when differentiating.

The constant added nothing to the growth of x, and does not enter into the differential coefficient. So any constant (even if represented with a letter) will disappear when we differentiate.

Differentiating products

The following first principles example is for establishing the product rule.

$$let \space y = (x^2 + c) \times (ax^4 + b)$$

You could multiply out the functions and differentiate the result:

$$y = ax^6 + bx^2 + acx^4 + bc$$

$$\frac{dy}{dx} = 6ax^5 + 2bx + 4acx^3$$

Or we could make use of first principles to establish a rule that we can make use of:

Consider $y = u \times v$

Where $u$ and $v$ are functions of x; and x grows to $x + dx$; $y + dy$; $u + du$; and $v + dv$ - we shall have

$$y + dy = (u + du) \times (v + dv)$$

$$y + dy = u \cdot v + u \cdot dv + v \cdot dv + du \cdot dv$$

$du \cdot dv$ is a small quantity of the second order of smallness, and can be disregarded. Leaving:

$$y + dy = u \cdot v + u \cdot dv + v \cdot du$$

Then subtracting the original ($y = u \cdot v$) we have left:

$$dy = u \cdot dv + v \cdot du$$

And dividing through by dx, we get:

$$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$$

Therefore this shows us our instructions when dealing with differentiating products. ( The product rule )

To differentiate the product of two functions, multiply each function by the differential coefficient of the other, and add together the two products obtained.

Essentially - treat $u$ as a constant while you differentiate $v$; then treat $v$ as a constant while you differentiate $u$; and the whole differential coefficient $\frac{dy}{dx}$ will be the sum of these two treatments.

Differentiating quotients

Lastly we have to differentiate quotients .

$$Let \space y = \frac{bx^5 + c}{x^2 + a}$$

In such a case it is no use to try to to work out the division beforehand, because $x^2 + a$ will not divide into $bx^5 + c$, neither have a common factor. So we have to go back to first principles to find a rule.

So we’ll put $y = \frac{u}{v}$

Where $u$ and $v$ are functions of the independent variable $x$.

Then when $y + dy$; $x + dx$; $v + dv$; and $u + du$ so that:

$$y + dy = \frac{u + du}{v + dv}$$

After performing algebraic division (omitted for brevity), you would get:

$$y + dy = \frac{u}{v} + \frac{v \cdot du - u \cdot dv}{v^2}$$

Now, subtract the original $y = \frac{u}{v}$, and we have the quotient rule as follows:

$$dy = \frac{v \cdot du - u \cdot dv}{v^2}$$


$$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$

Thank you for reading!