DIFFERENTIATING FROM THE FIRST PRINCIPLES

J.O. EMMANUEL

To differentiate using the first principle, Here are the steps.

First, you make y increases by dy and x increase by dx

Secondly, you subtract y from both sides

Thirdly, you divide both sides by dx

Lastly, you apply the limit expression as dx tends to approach 0

The next three examples illustrate this;

Example 1

Given that y=x, Find $\frac{dy}{dx}$

Solution:

$y=x$

Let y increase by $dy$ and x increase by $dx$.

$$y+dy=x+dx$$

Now, we subtract $-y$ from both sides

$$dy=x+dx-y$$

Since $y=x$

$$dy=x+dx-(x)$$

$$dy=dx$$

Now, let divide both sides $dx$

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

$$\frac{dy}{dx}=1$$

Example 2

From the first principle, differentiate  $y=x^2+4$

Solution:

Let y increase by $dy$ and x increase by $dx$.

$$(y+dy)=(x+dx)^2+4$$

$$y+dy=x(x+dx)+dx(x+dx)+4$$

$$dy=x^2+2xdx+(dx)^2+4-y$$

If $y=x^2+4$, then

$$dy=x^2+2xdx+(dx)^2+4-(x^2+4)$$

Now, let distribute the negative sign

$$dy=x^2+2xdx+(dx)^2+4-x^2-4$$

$$dy=2xdx+(dx)^2$$

$$\frac{dy}{dx}=\frac{dx(2x+dx)}{dx}$$

Applying limit expression

$$\frac{dy}{dx}=\lim_{dx \to 0} 2x+(0)$$

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

Example 3

Using first principle, find the $\frac{dy}{dx}$ of $y=x^2+x+4$

Solution:

$$y=x^2+x+4$$

Let y increase by $dy$ and x increase by $dx$.

$$y+dy=(x+dx)^2+(x+dx)+4$$

Expanding the expression

$$y+dy=x^2+2xdx+(dx)^2+dx+x+4$$

Now, let subtract y from both sides

$$dy=x^2+2xdx+(dx)^2+dx+x+4-y$$

If $y=x^2+x+4$, then

$$dy=x^2+2xdx+(dx)^2+dx+x+4-(x^2+x+4)$$

Now, let's distribute the negative sign

$$dy=x^2+2xdx+(dx)^2+dx+x+4-x^2-x-4$$

$$dy=2xdx+(dx)^2+dx$$

Dividing both sides by $dx$

$$\frac{dy}{dx}=2x+dx+1$$

Now, let's apply the limit expression

$$\frac{dy}{dx}=\lim_{dx \to 0}2x+0+1$$

$$\frac{dy}{dx}=2x+1$$

$$\frac{dy}{dx}=2x+1$$

That will be all for now. In the next post, we will be learning how to find the derivative of a function.

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