Delta Process for Derivatives

Delta Process is one of the many ways to determine the derivative of a given function. Let's consider a function f(x) or y shown below on which we will apply delta process to find its derivative.
\[y=\frac{2}{x^2} \tag1\]
This method finds the change in $y$ with respect to the change in $x$ and we take the limit of this slope as independent variable approaches zero. To find the change in $x$ and $y$, we add add little perturbation to these parameters as shown below:
\[y+\Delta y = \frac{2}{(x+\Delta x)^2} \tag2\]
Subtract $y$ from both sides to isolate $\Delta y$ on the left.
\[\Delta y = \frac{2}{(x+\Delta x)^2} - y \tag3\]
In Equation 3, substitute the original definition for $y$ shown in Equation 1.
\[\Delta y = \frac{2}{(x+\Delta x)^2} - \frac{2}{x^2} \tag4\]
Now combine these two fractions by making their denominators common.
\[\Delta y = \frac{2x^2 - 2(x+\Delta x)^2}{(x+\Delta x)^2 x^2} \tag5\]
Simplify the right side by expanding the squared term in the numerator. You don't necessarily have to simply the denominator because there is no terms that could be canceled.
\[(x+\Delta x)^2 = x^2 +2x\Delta x +(\Delta x)^2 \tag6\]
Substitute Eq. 6 in the numerator of Eq. 5. After substitution Eq. 5 becomes:
\[\Delta y = \frac{2x^2 - 2x^2 -4x\Delta x - (\Delta x)^2}{(x+\Delta x)^2 x^2} \tag7\]
\[\Delta y = \frac{-4x\Delta x - (\Delta x)^2}{(x+\Delta x)^2 x^2} \tag8\]
Divide both side by $\Delta x$
\[\frac{\Delta y}{\Delta x} = \frac{\Bigg[\frac{-4x\Delta x - 2\Delta x}{(x+\Delta x)^2 x^2}\Bigg]}{\Delta x} \tag9\]
\[\frac{\Delta y}{\Delta x} = \Bigg[\frac{-4x\Delta x - 2\Delta x}{(x+\Delta x)^2 x^2}\Bigg]\frac{1}{\Delta x} = \frac{-4x-\Delta x}{(x+\Delta x)^2 x^2}\tag{10}\]
$\frac{\Delta y}{\Delta x}$ represents the slope of the original function that we are trying to differentiate. Now, we have to take the limit of this slope as the $\Delta x$ approaches zero.
\[\lim_{\Delta x\to\infty}\frac{\Delta y}{\Delta x} = \lim_{\Delta x\to\infty}\frac{-4x-\Delta x}{(x+\Delta x)^2 x^2} \tag{11}\]
The left side of the equation can be represented by $y'$, the conventional symbol for the derivative, and the right side becomes:
\[y' = \frac{-4x}{x^4} \tag{12} \]
Simplify further:
\[y' = \frac{-4}{x^3} \tag{13}\]
Equation 13 is the derivative of the original function that we have in Equation 1 above. This is how we use Delta Process to differentiate a function. Let me know in the comments below if you find something hard to understand and I will try to answer.