# The Derivative by Definition

## Key Questions

• First you have to calculate the derivative of the function.

$f \left(x\right) = {x}^{3}$

$f ' \left(x\right) = 3 {x}^{2}$

Then if we want to find the derivative of $f \left(x\right)$ when $x = 4$ then we substitute that value into $f ' \left(x\right)$.

$f ' \left(4\right) = 3 {\left(4\right)}^{2} = 3 \cdot 16 = 48$

• The formal definition of derivative of a function $y = f \left(x\right)$ is:
$y ' = {\lim}_{\Delta x \to 0} \frac{f \left(x + \Delta x\right) - f \left(x\right)}{\Delta x}$

The meaning of this is best understood observing the following diagram: The secant PQ represents the mean rate of change $\frac{\Delta y}{\Delta x}$ of your function in the interval between $x$ and $x + \Delta x$.

If you want the rate of change, say, at P you "move" point Q (and the secant with it) to meet point P as in: In doing so you must reduce $\Delta x$. If $\Delta x \to 0$ you'll get the tangent in P whose inclination will give the inclination at P and thus the derivative at P. Hope it helps!

We use quotient rule as described below to differentiate algebraic fractions or any other function written as quotient or fraction of two functions or expressions

#### Explanation:

When we are given a fraction say $f \left(x\right) = \frac{3 - 2 x - {x}^{2}}{{x}^{2} - 1}$. This comprises of two fractions - say one $g \left(x\right) = 3 - 2 x - {x}^{2}$ in numerator and the other $h \left(x\right) = {x}^{2} - 1$, in the denominator. Here we use quotient rule as described below.

Quotient rule states if $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$

then $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\frac{\mathrm{dg}}{\mathrm{dx}} \times h \left(x\right) - \frac{\mathrm{dh}}{\mathrm{dx}} \times g \left(x\right)}{h \left(x\right)} ^ 2$

Here $g \left(x\right) = 3 - 2 x - {x}^{2}$ and hence $\frac{\mathrm{dg}}{\mathrm{dx}} = - 2 - 2 x$ and as $h \left(x\right) = {x}^{2} - 1$, we have $\frac{\mathrm{dh}}{\mathrm{dx}} = 2 x$ and hence

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\left(- 2 - 2 x\right) \times \left({x}^{2} - 1\right) - 2 x \times \left(3 - 2 x - {x}^{2}\right)}{{x}^{2} - 1} ^ 2$

= $\frac{- 2 {x}^{3} - 2 {x}^{2} + 2 x + 2 - 6 x + 4 {x}^{2} + 2 {x}^{3}}{{x}^{2} - 1} ^ 2$

= $\frac{2 {x}^{2} - 4 x + 2}{{x}^{2} - 1} ^ 2$

or $\frac{2 {\left(x - 1\right)}^{2}}{{x}^{2} - 1} ^ 2$

= $\frac{2}{x + 1} ^ 2$

Observe that $\frac{3 - 2 x - {x}^{2}}{{x}^{2} - 1} = \frac{\left(1 - x\right) \left(3 + x\right)}{\left(x + 1\right) \left(x - 1\right)} = \frac{- 3 - x}{x + 1}$ and using quotient rule

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{- \left(x + 1\right) - \left(- 3 - x\right)}{x + 1} ^ 2 = \frac{2}{x + 1} ^ 2$