Question

Find the derivatives of the following using increment method.1. y= 3x² - 2x + 52.y = 6x² +10x - 3

Answer 1

Answer:

y'=6x-2

Explanation:

Given the function:

$y=3x^2-2x+5$

We want to find the derivative using the increment method.

First, replace y with y+△y and x with x+△x.

$\begin{gathered} y+\Delta y=3(x+\Delta x)^2-2(x+\Delta x)+5 \\ \text{Make }\Delta y\text{ the subject of the equation} \\ \Delta y=3(x+\Delta x)^2-2(x+\Delta x)+5-y \\ \text{ Replace y with the initial function given} \\ \Delta y=3(x+\Delta x)^2-2(x+\Delta x)+5-(3x^2-2x+5) \end{gathered}$

Next, simplify the right-side of the equation:

$\begin{gathered} \Delta y=3(x^2+2x\Delta x+(\Delta x)^2)-2x-2\Delta x+5-3x^2+2x-5 \\ \Delta y=3x^2+6x\Delta x+3(\Delta x)^2-2x-2\Delta x+5-3x^2+2x-5 \\ \Delta y=3x^2-3x^2+6x\Delta x+3(\Delta x)^2-2x+2x-2\Delta x+5-5 \\ \Delta y=6x\Delta x+3(\Delta x)^2-2\Delta x \end{gathered}$

Factor out △x in the right-side of the equation:

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

Divide both sides by △x:

$\begin{gathered} \frac{\Delta y}{\Delta x}=\frac{\Delta x(6x+3\Delta x-2)}{\Delta x} \\ \frac{\Delta y}{\Delta x}=6x-2+3\Delta x \end{gathered}$

Let △x tends to 0:

$\frac{\Delta y}{\Delta x}=6x-2$

The derivative of y is 6x-2.