General rule for differentiation

From the definition of a derivative it is seen that the process of differentiating a function $y = f(x)$ consists in taking the following distinct steps:

General rule for differentiating^4.5:

• FIRST STEP. In the function replace $x$ by $x + \Delta\, x$, giving a new value of the function, $y + \Delta\, y$.

• SECOND STEP. Subtract the given value of the function from the new value in order to find $\Delta\, y$ (the increment of the function).

• THIRD STEP. Divide the remainder $\Delta\, y$ (the increment of the function) by $\Delta\, x$ (the increment of the independent variable).

• FOURTH STEP. Find the limit of this quotient, when $\Delta\, x$ (the increment of the independent variable) varies and approaches the limit zero. This is the derivative required.

The student should become thoroughly familiar with this rule by applying the process to a large number of examples. Three such examples will now be worked out in detail.

Example 4.7.1   Differentiate $3x^2 + 5$.

Solution. Applying the successive steps in the General Rule, we get, after placing

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

First step.

$$y + \Delta\, y = 3(x + \Delta\, x)2 + 5 = 3x^2 + 6x \cdot \Delta x + 3(\Delta x)^2 + 5.$$

Second step.

$$\begin{array}{ll} y + \Delta\, y & = 3x^2 + 6 x \cdot \Delta... ... \Delta\, y & = 6x \cdot \Delta x + 3(\Delta x)^2. \end{array}$$

Third step. $\frac{\Delta y}{\Delta x} = 6x + 3 \cdot \Delta x.$

Fourth step. $\frac{dy}{dx} = 6x$. We may also write this

$$\frac{d}{dx} (3x^2 + 5) = 6x.$$

Here's how to use SAGE to verify this (for simplicity, we set $h=\Delta x$):

[fontsize=\scriptsize,fontfamily=courier,fontshape=tt,frame=single,label=\sage]

sage: x = var("x")
sage: h = var("h")
sage: f(x) = 3*x^2 + 5
sage: Deltay = f(x+h)-f(x)
sage: (Deltay/h).expand()
6*x + 3*h
sage: limit((f(x+h)-f(x))/h,h=0)
6*x
sage: diff(f(x),x)
6*x

Example 4.7.2   Differentiate $x^3 - 2x + 7$.

Solution. Place $y = x^3 - 2x + 7$.

First step.

$$\begin{array}{ll} y + \Delta y & = (x + \Delta x)3 - 2(x + ... ...lta x)^2 + (\Delta x)^3 - 2x - 2 \cdot \Delta x + 7 \end{array}$$

Second step.

$$\begin{array}{ll} y + \Delta y & = x^3 + 3x^2 \cdot \Delta x... ...dot (\Delta x)^2 + (\Delta x)^3 - 2 \cdot \Delta x \end{array}$$

Third step. $\frac{\Delta y}{\Delta x} = 3x^2 + 3x \cdot \Delta x + (\Delta x)^2 - 2$.

Fourth step. $\frac{dy}{dx} = 3x^2 - 2$. Or,

$$\frac{d}{dx} (x^3 - 2x + 7) = 3x^2 - 2.$$

Example 4.7.3   Differentiate $\frac{c}{x^2}$.

Solution. Place $y = \frac{c}{x^2}$.

First step. $y + \Delta y = \frac{c}{(x + \Delta x)^2}$.

Second step.

$$\begin{array}{ll} y + \Delta y & = \frac{c}{(x + \Delta x)^... ...cdot \Delta x(2x + \Delta x)}{x^2(x + \Delta x)^2}. \end{array}$$

Third step. $\frac{\Delta y}{\Delta x} = -c \cdot \frac{2x + \Delta x}{x^2(x + \Delta x)^2}$.

Fourth step. $\frac{dy}{dx} = -c \cdot \frac{2x}{x^2(x)^2} = -\frac{2c}{x^3}$. Or, $\frac{d}{dx} \left ( \frac{c}{x^2} \right ) = \frac{-2c}{x^3}$.