If we have a real valued function then the derivative of this function will measure the rate of change of the dependent variable or the quantity that is determined by the independent variable or another quantity.

Also, the derivative of a function in single variable gives us the function of the slope of the line that is tangent to the curve of the given function.

The derivative of a function $‘y’$ is represented as $\frac{dy}{dx}$.

Let $g$ be a real valued function and the tangent line at a point $(a, g (a))$. The derivative of g at a is the slope of the tangent line. The slope of tangent line will be very near to the slope of line that is passing through the point $(a, g (a))$ and a point near say $(a + k, g (a + k))$.

So $m$ = $\frac{(g (a + k) – g (a))}{k}$

As $k$ $\rightarrow$ 0 we take limits,

$g’(a)$ = $lim_{k \rightarrow 0}$ $[\frac{(g (a + k) – g (a))}{k}]$

We say when this limits does exist then g is differentiable at point ‘a’.

The derivative equivalently will also satisfy the property that:

$lim_{h \rightarrow 0}$ $[(\frac{g (a + k) – g (a) – g’(a) . k)}{k}]$ = 0

We can take derivative further of a derivative function. These derivatives of derivatives are known as higher order derivatives and are named as their step.

The point where the second derivative of any function changes sign is known as the inflection point.

Below are some rules help to find the derivatives of the various functions:

**1)** If $g(x)$ = $x^m$ then $g’(x)$ = $m x^{m - 1}$

$\frac{d}{dx}$ $(a^x)$ = ln (a) $a^x$

$\frac{d}{dx}$ $(ln x)$ = $\frac{1}{x}$, x > 0

$\frac{d}{dx}$ $log_a (x)$ = $\frac{1}{(x\ ln\ a)}$

$\frac{d}{dx}$ $(cos x)$ = -$sin x$

$\frac{d}{dx}$ $(tan x)$ = $sec^2 x$ = 1 + $tan^2 x$

$\frac{d}{dx}$ $(cosec x)$ = -$cosec x cot x$

$\frac{d}{dx}$ $(sec x)$ = $sec x tan x$

$\frac{d}{dx}$ $(cot x)$ = - $cosec^2 x$

Let $f(x)$ and $g(x)$ be two functions. Then, $(f(x) g(x))’$ = $f(x) g’(x)$ + $g(x) f’(x)$

($\frac{f(x)}{g(x)}$)’ = $\frac{f’(x) g(x) - f(x) g’(x)}{(g(x))^2}$, $g \neq 0$

$(f (g( x)))’$ = $f’ (g (x))$ . $g’(x)$

Also if $y = y (u)$ then, $\frac{dy}{dx}$ = $\frac{dy}{du}$ * $\frac{du}{dx}$

We make use of product rule.

$\frac{d}{dx}$ [$(x + 1) (x – 3)$] = $(x + 1) (1)$ + $(x – 3) (1)$ = $x + 1$ + $x – 3$ = $2x – 2$

$\frac{d}{dx}$ [$\frac{(x^3 – 3 x^2)}{x^2}$] = $\frac{x^2 (3x^2 – 6 x)- 2 x (x^3 – 3 x^2)}{x^4}$

= $\frac{(3 x^4 – 6 x^3 – 2 x^4 + 6 x^3)}{x^4}$

= $\frac{x^4}{x^4}$

= 1.

$\frac {dy}{dx}$ = $\frac {d}{dx}$( $\sqrt{x^2+2}$)

$\frac {dy}{dx}$ = $\frac {1}{2}$( $(x^2+2)^{\frac{-1}{2}} $)$\frac {d}{dx}$( ${x^2+2}$)

$\frac {dy}{dx}$ = $\frac {1}{2 \sqrt{x^2+2}}$ $\times$ 2x

$\frac {dy}{dx}$ = $\frac {x}{ \sqrt{x^2+2}}$