Derivative (Increment Method)

The Derivative using Increment Method

The increment method is a technique used in calculus to derive mathematical formulas that describe the behavior of a function as its input variable changes. The technique is particularly useful in cases where the function is not easily differentiable or when the function is defined only in terms of discrete values.

The basic idea behind the increment method is to calculate the difference between two values of a function as the input variable changes by a small amount, which is referred to as the increment. By taking the limit of this difference as the increment approaches zero, we can obtain the derivative of the function at a particular point.

The advantage of using the increment method is that it allows us to approximate the behavior of a function using only the information that is readily available. For example, if we know the values of a function at discrete points, we can use the increment method to derive a formula that describes the rate at which the function changes between those points.

Another advantage of the increment method is that it can be used to derive formulas for functions that are not continuous or differentiable at certain points. For example, if a function has a jump discontinuity at a particular point, we can use the increment method to derive a formula that describes the behavior of the function on either side of the point.

This is the basic concept of changes that happened mathematically. Speaking of derivative, it's just saying we're dismantling a thing just to see the parts of it. Usually, teachers' illustrative example is a loaf of bread sliced just to see what was inside. Actually, there is a lot. Anything that approaches its end is an example of Differential Calculus nature. Even drinking water, notice that the 100% is approaching zero when the glass is empty. But the clearest example is anything visual. We are presenting the increment method on this page. This is a bit tricky, so just take it easy.

This type of mathematics requires exposure to Algebra. Since the operations are purely Algebraic.

We prepared the basic mathematical examples below. We tried highlighting the parts so that it won't ur your vision. There are three illustrative examples with aided discussions and added-by-example problems with solutions for further practice. If you notice there's a typographical error, feel free to let us know by telling us here. Let's start!

Illustrative Examples

Find the first derivative using the increment method.

1.) ${\color{Blue} y}=3{\color{Red} x}^{2}-{\color{Red} x}+1$

${\color{Blue} y+\Delta y}=3{\color{Red} (x+\Delta x)}^{2}-{\color{Red} (x+\Delta x)}+1$

${\color{Blue} y+\Delta y}=3{\color{Red} (x^{2}+2x\Delta x+\Delta x^{})}-{\color{Red} (x+\Delta x)}+1$

${\color{Blue} y+\Delta y}=3{\color{Red} x^{2}}+{\color{Red} 6x\Delta x}+3{\color{Red} (\Delta x)}^{2}-{\color{Red} x}-{\color{Red} \Delta x}+1$

${\color{Blue} \Delta y}=3{\color{Red} x^{2}}+{\color{Red} 6x\Delta x}+{\color{Red} 3(\Delta x)^{2}}-{\color{Red} x}-{\color{Red} \Delta x}+1-{\color{Blue} y}$

   ${\color{Blue} \Delta y}=3{\color{Red} x^{2}}+{\color{Red} 6x\Delta x}+3({\color{Red} \Delta x})^{2}-{\color{Red} x}-{\color{Red} \Delta x}+1-({\color{Blue} 3x^{2}-x+1})$

${\color{Blue} \Delta y}=3{\color{Red} x^{2}}+{\color{Red} 6x\Delta x}+3({\color{Red} \Delta x})^{2}-{\color{Red} x}-{\color{Red} \Delta x}+1-\color{Blue} 3x^{2}+x-1$

${\color{Blue} \Delta y}={\color{DarkGreen} {3{ x^{2}}}}+{\color{Red} 6x\Delta x}+3({\color{Red} \Delta x})^{2}-{\color{DarkGreen} x}-{\color{Red} \Delta x}+1-{\color{DarkGreen} 3x^{2}}+{\color{DarkGreen} x}-1$ , cancel all dark greens

   ${\color{Blue} \Delta y}={\color{Red} 6x\Delta x}-{\color{Red} \Delta x}+{\color{Red} 3(\Delta x)^{2}}$

   $\frac{{\color{Blue} \Delta y}}{{\color{Red} \Delta x}}=\frac{{\color{Red} \Delta x}(6x-1+3\Delta x)}{\color{Red} \Delta x}$ , cross out ${\color{Red} \Delta x}$

   $\frac{{\color{Blue} \Delta y}}{{\color{Red} \Delta x}}=(6x-1+3\Delta x)$ , ${\color{Red} \Delta x}\rightarrow 0$

   $\frac{{\color{Blue} \Delta y}}{{\color{Red} \Delta x}}=(6x-1+3(0))$

   $\frac{{\color{Blue} \Delta y}}{{\color{Red} \Delta x}}=6x-1$ , answer

2.) $y={\color{DarkRed} 3x^{2}-4x+7}$

${\color{Red} y}+\Delta y=3(x+\Delta x)^{2}-4(x+\Delta x)+7$

${\color{Red} y}+\Delta y=3[{x^{2}+2x\Delta x+(\Delta x)^{2}}]-(4x+\Delta x)+7$

$\Delta y=3x^{2}+6x\Delta x+3(\Delta x)^{2}-4x-4\Delta x+7-{\color{DarkRed} 3x^{2}+4x-7}$ , remember y as given eq,n?

$\Delta y={\color{DarkGreen} 3x^{2}}+6x\Delta x+3(\Delta x)^{2}{\color{DarkGreen} -4x}-4\Delta x{\color{DarkGreen} +7}{\color{DarkGreen} -3x^{2}}{\color{DarkGreen} +4x}{\color{DarkGreen} -7}$ , compensate similar terms

$\Delta y=6x{\color{Red} \Delta} x+3{\color{Red} \Delta} x^{2}-4{\color{Red} \Delta} x$

$\frac{\Delta y}{{\color{Red} \Delta x}}=\frac{{\color{Red} \Delta x}(6x+3\Delta x-4)}{{\color{Red} \Delta x}}$

$\frac{\Delta y}{{\color{Red} \Delta x}}=6x+3{\color{Red} \Delta x}-4$ , ${\color{Red} \Delta x}\rightarrow 0$

$\frac{\Delta y}{{\color{Red} \Delta x}}=6x+3{\color{Red}(0 )}-4$

$\frac{\Delta y}{{\color{Red} \Delta x}}=6x-4$ , answer

3.) $y={\color{DarkRed} 1+3x-2x^{2}}$

$y+\Delta y=1+3(x+\Delta x)-2(x+\Delta x )^{2}$

$y+\Delta y=1+3(x+{\color{Red} \Delta x})-2[x^{2}+2x{\color{Red} \Delta x} +({\color{Red} \Delta x})^{2}]$

$\Delta y=1+3x+3{\color{Red} \Delta x}-2x^{2}-4x{\color{Red} \Delta x}-2({\color{Red} \Delta x})^{2}$

$\Delta y=1+3x+3{\color{Red} \Delta x}-2x^{2}-4x{\color{Red} \Delta x}-2({\color{Red} \Delta x})^{2}-y$ , this y is the given eq'n.

$\Delta y=1+3x+3{\color{Red} \Delta x}-2x^{2}-4x{\color{Red} \Delta x}-2({\color{Red} \Delta x})^{2}{\color{DarkRed} -1-3x+2x^{2}}$ , substituting the value of y

$\Delta y={\color{DarkGreen} 1}{\color{DarkGreen} +3x}+3{\color{Red} \Delta x}{\color{DarkGreen} -2x^{2}}-4x{\color{Red} \Delta x}-2({\color{Red} \Delta x})^{2}{\color{DarkGreen} -1-3x+2x^{2}}$ , compensate dark green color

$\Delta y={3{\color{Red} \Delta x}}+4x{\color{Red} \Delta x} -2({\color{Red} \Delta x})^{2}$

$\Delta y=3{\color{Red} \Delta x}-4x{\color{Red} \Delta x}-2({\color{Red} \Delta x})^{2}$

$\frac{\Delta y}{{\color{Red} \Delta x}}=\frac{{\color{Red} \Delta x}(3-4x-2{\color{Red} \Delta x})}{\color{Red} \Delta x}$ , cross out ${\color{Red} \Delta x}$

$\frac{\Delta y}{{\color{Red} \Delta x}}=3-4x-2{\color{Red} \Delta x}$ , ${\color{Red} \Delta x}\rightarrow 0$

$\frac{\Delta y}{{\color{Red} \Delta x}}=3-4x-2({\color{Red} 0})$

$\frac{\Delta y}{{\color{Red} \Delta x}}=3-4x$ , answer

Example problems with solutions:

Find the first derivative using the increment method.
These problems can be found in the textbook "Differential and Integral Calculus by Love and Rainville"

$1.) y=4x^{2}-3x-2$

$y+\Delta y=4(x+\Delta x)^{2}-3(x+\Delta x)-2$

$y+\Delta y=4[(x^{2}+2x\Delta x+(\Delta x)^{2}]-3(x+\Delta x)-2$

$\Delta y=4[(x^{2}+2x\Delta x+(\Delta x)^{2}]-3(x+\Delta x)-2-y$ , y from the given eq'n

$\Delta y=4[(x^{2}+2x\Delta x+(\Delta x)^{2}]-3(x+\Delta x)-2-(4x^{2}-3x-2)$ , substitute the value of
y

$\Delta y=4x^{2}+8x\Delta x+4\Delta x^{2}-3x-3\Delta x-2{\color{DarkRed} -4x^{2}+3x+2}$ , removing parenthesis and collecting/compensating for similar terms.

$\Delta y={\color{DarkRed} 4x^{2}}+8x\Delta x+4\Delta x^{2}{\color{DarkRed} -3x}-3\Delta x{\color{DarkRed} -2}{\color{DarkRed} -4x^{2}+3x+2}$ , compensate all in red

$\Delta y=8x\Delta x+4\Delta x^{2}-3\Delta x\color{DarkRed}$

$\frac{\Delta y}{\Delta x}=\frac{\Delta x(8x+4\Delta x-3)}{\Delta x}$ , cross out ${\color{Red} \Delta x}$

$\frac{\Delta y}{\Delta x}=8x+4\Delta x-3$ , ${\color{Red} \Delta x}\rightarrow 0$

$\frac{\Delta y}{\Delta x}=8x+4(0) -3$

$\frac{\Delta y}{\Delta x}=8x-3$ , answer

$2.)y=3+4x-x^{2}$

$y+\Delta y=3+4(x+\Delta x)-(x+\Delta x)^{2}$

$y+\Delta y=3+4(x+\Delta x)-[x^{2}+2x\Delta x+(\Delta x)^{2}]$

$\Delta y=3+4(x+\Delta x)-[x^{2}+2x\Delta x+(\Delta x)^{2}]-y$

$\Delta y=3+4(x+\Delta x)-[x^{2}+2x\Delta x+(\Delta x)^{2}]-(3+4x-x^{2})$

$\Delta y=3+4x+4\Delta x-x^{2}-2x\Delta x-(\Delta x)^{2}-3-4x+x^{2}$

$\Delta y={\color{DarkGreen} 3}{\color{DarkGreen} +4x}+4\Delta x{\color{DarkGreen} -x^{2}}-2x\Delta x-(\Delta x)^{2}{\color{DarkGreen} -3-4x+x^{2}}$

$\Delta y=4\Delta x-2x\Delta x-(\Delta x)^{2}$

$\frac{\Delta y}{\Delta x}=\frac{\Delta x(4-2x-\Delta x)}{\Delta x}$

$\frac{\Delta y}{\Delta x}=4-2x-\Delta x$ , ${\color{Red} \Delta x}\rightarrow 0$

$\frac{\Delta y}{\Delta x}=4-2x-0$

$\frac{\Delta y}{\Delta x}=4-2x$ , answer

$3.) y=x^{3}-5x+2$

$y+\Delta y=(x+\Delta x)^{3}-5(x+\Delta x)+2$

$\Delta y=x^{3}+3x^{2}\Delta x+3x(\Delta x)^{2}+(\Delta x)^{3}-5(x+\Delta x)+2-y$

$\Delta y=x^{3}+3x^{2}\Delta x+3x(\Delta x)^{2}+(\Delta x)^{3}-5(x+\Delta x)+2-(x^{3}-5x+2)$

$\Delta y=x^{3}+3x^{2}\Delta x+3x(\Delta x)^{2}+(\Delta x)^{3}-5x-5\Delta x+2-x^{3}+5x-2$

$\Delta y={\color{DarkGreen} x^{3}}+3x^{2}\Delta x+3x(\Delta x)^{2}+(\Delta x)^{3}{\color{DarkGreen} -5x}-5\Delta x{\color{DarkGreen} +2}{\color{DarkGreen} -x^{3}+5x-2}$

$\Delta y=3x^{2}\Delta x+3x(\Delta x )^{2}+(\Delta x )^{3}-5\Delta x$

$\frac{\Delta y}{\Delta x}=\frac{\Delta x[(3x^{2}+3x\Delta x)+(\Delta x)^{2}-5]}{\Delta x}$

$\frac{\Delta y}{\Delta x}={3x^{2}+3x\Delta x+(\Delta x)^{2}-5}$ , ${\color{Red} \Delta x}\rightarrow 0$

$\frac{\Delta y}{\Delta x}={3x^{2}+3x(0)+(0)^{2}-5}$

$\frac{\Delta y}{\Delta x}={3x^{2}-5}$ , answer

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

$y+\Delta y=(x+\Delta x)^{3}-2(x+\Delta x)^{2}+6$

$y+\Delta y=[x^{3}+3x^{2}\Delta x+3x(\Delta x)^{2}+(\Delta x)^{3}]-2[x^{2}+2x \Delta x+(\Delta x)^{2}]+6$

$\Delta y=[x^{3}+3x^{2}\Delta x+3x(\Delta x)^{2}+(\Delta x)^{3}]-2[x^{2}+2x \Delta x+(\Delta x)^{2}]+6-y$

$\Delta y=[x^{3}+3x^{2}\Delta x+3x(\Delta x)^{2}+(\Delta x)^{3}]-2[x^{2}+2x \Delta x+(\Delta x)^{2}]+6-(x^{3}-2x^{2}+6)$

$\Delta y=x^{3}+3x^{2}\Delta x+3x(\Delta x)^{2}+(\Delta x)^{3}-2x^{2}-2x \Delta x-(\Delta x)^{2}]+6-x^{3}+2x^{2}-6$

$\Delta y={\color{DarkGreen} x^{3}}+3x^{2}\Delta x+3x(\Delta x)^{2}+(\Delta x)^{3}{\color{DarkGreen} -2x^{2}}-2x \Delta x-(\Delta x)^{2}]{\color{DarkGreen} +6}{\color{DarkGreen} -x^{3}+2x^{2}-6}$

$\Delta y=3x^{2}\Delta x+3x(\Delta x)^{2}+(\Delta x)^{3}-4x \Delta x-(\Delta x)^{2}$

$\frac{\Delta y}{\Delta x}=\frac{[\Delta x(3x^{2}+3x\Delta x+(\Delta x)^{2}-4x-\Delta x]}{\Delta x}$

$\frac{\Delta y}{\Delta x}=3x^{2}+3x\Delta x+(\Delta x)^{2}-4x-\Delta x$ , ${\color{Red} \Delta x}\rightarrow 0$

$\frac{\Delta y}{\Delta x}=3x^{2}+3x(0)+(0)^{2}-4x-0$

$\frac{\Delta y}{\Delta x}=3x^{2}-4x$ , answer

$5.) y=(x-2)^{2}$

$y=x^{2}-2x+4$

$y+\Delta y=(x+\Delta x)^{2}-2(x+\Delta x)+4$

$y+\Delta y=[x^{2}+2x\Delta x+(\Delta x)^{2}]-2x-2\Delta x+4$

$\Delta y=[x^{2}+2x\Delta x+(\Delta x)^{2}]-2x-2\Delta x+4 -y$

$\Delta y=[x^{2}+2x\Delta x+(\Delta x)^{2}]-2x-2\Delta x+4 -(x^{2}-2x+4)$

$\Delta y=x^{2}+2x\Delta x+(\Delta x)^{2}-2x-2\Delta x+4 -x^{2}+2x-4$

$\Delta y={\color{DarkGreen} x^{2}}+2x\Delta x+(\Delta x)^{2}{\color{DarkGreen} -2x}-2\Delta x{\color{DarkGreen} +4} {\color{DarkGreen} -x^{2}+2x-4}$

$\Delta y=2x\Delta x+(\Delta x)^{2}-2\Delta x$

$\frac{\Delta y}{\Delta x}=\frac{\Delta x[2x+\Delta x-2]}{\Delta x}$ ,

$\frac{\Delta y}{\Delta x}={2x+\Delta x-2}$ , ${\color{Red} \Delta x}\rightarrow 0$

$\frac{\Delta y}{\Delta x}={2x-2}$ , answer

Seatwork with Partial Solutions:

Find the first derivative using the increment method.

$1.)y=(2x-1)^{2}=4x^{2}-4x+1$

$y+\Delta y=4(x+\Delta x)^{2}-4(x+\Delta x)+1$

$y+\Delta y=4[x^{2}+2x\Delta x+(\Delta)^{2} ]-4(x+\Delta x)+1$

$\Delta y=4[x^{2}+2x\Delta x+(\Delta)^{2} ]-4(x+\Delta x)+1-y$

Click to continue

$2.)x=\frac{1}{2}t^{2}-t+2$

$x+\Delta x=\frac{1}{2}(t+\Delta t)^{2}-(t+\Delta t)+2$

$x+\Delta x=\frac{1}{2}[t^{2}+2t\Delta t+(\Delta t)^{2}]-(t+\Delta t)+2$

$\Delta x=\frac{1}{2}[t^{2}+2t\Delta t+(\Delta t)^{2}]-(t+\Delta t)+2-x$

Click to continue

$3.)y=3+4x-x^{2}$

$y+\Delta y=3+4(x+\Delta x)-(x+\Delta x)^{2}$

$y+\Delta y=3+4(x+\Delta x)-[x^{2}+2x\Delta x+(\Delta x)^{2}]$

$\Delta y=3+4(x+\Delta x)-[x^{2}+2x\Delta x+(\Delta x)^{2}]-y$

Click to continue

Simple Quiz:

Find the first derivative using the increment method.

$1.) y=x^{2}-4x-10$

$2.) y=7-5x-2x^{2}$

$3.) y=2x^{3}+7x-4$

$4.) y=\frac{2}{3}x^{3}+7x-4$

$5.) y=(3x^{2}-4)^{2}$

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