In Calculus, a fundamental concept of representing the instantaneous rate of change of a function with respect to one of its variable is called Derivative. The first derivative of any function is a function whose values can be interpreted as slopes of tangent lines to the graph of the original function at a given point. This process of finding a derivative of a function is called as Differentiation.
Derivatives can also be defined geometrically.

Differentiation

The derivative of a function is a function whose values are interpreted as slopes of tangent lines. Let us find the slope of the tangent line to a graph at the point P. The slope can be approximated by drawing a line through the point P and another point nearby, and then finding the slope of that line, called a secant line. The slope of a line define by a function is determined using the following formula (m represents slope): m = $\frac{rise}{run}$ = $\frac{\Delta{y}}{\Delta{x}}$
Let P = (x$_1$, y$_1$) and Q = (x$_2$, y$_2$). Let x$_2$ = x$_1$ + $\Delta{x}$ and y$_2$ = y$_1$ + $\Delta{y}$ and $\Delta{y}$ = f(x + $\Delta{x}$) – f(x).

Then the slope of the line is given by $\frac{y_2 – y_1}{x_2 – x_1}$ = $\frac{\Delta{y}}{\Delta{x}}$ = $\frac{f(x + \Delta{x}) – f(x)}{\Delta{x}}$

We chose any arbitrary interval to be $\Delta$x. The smaller $\Delta{x}$ is, the more accurate this approximation is. So we decrease the size of $\Delta$x as much as possible by taking the limits as $\Delta{x}$ approaches zero. Assuming the limit exists, we find the exact slope of the tangent line to the curve at the given point and this values is the Derivative given by $\frac{df}{dx}$ = $\lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}$This leads to expressing the definition of derivative in three different ways.

$\frac{df}{dx}$ =

$\lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} = \lim_{\Delta h \rightarrow 0} \frac{f(x + h) - f(x)}{h} = \lim_{\Delta x \rightarrow 0} \frac{\Delta y}{\Delta x}$

Derivative of a function y = f(x) is represented as f'(x) or $\frac{dy}{dx}$ or y' or $\frac{d}{dx}$ [f(x)] or Df(x).

The basic rules of Differentiation of functions in calculus are shown below along with examples:

1) Derivative of a constant function.
The derivative of a function f(x) = c where c is a constant is given by f '(x) = 0.
Example: If f(x) = 7, then f'(x) = 0.


2) Derivative of a power function (power rule).
The derivative of a function f(x) = $x^n$ where n is a constant real number is given by f'(x) = n $x^{n-1}$.
Example: If f(x) = $x^4$, then f'(x) = $4x^3$.
If f(x) = $x^{-5}$ then f’(x) = -5$x^{(-5-1)}$ = -5$x^{-6}$ = -$\frac{5}{x^6}$.


3) Derivative of a function multiplied by a constant.
The derivative of a function f(x) = c g(x) is given by f'(x) = cg'(x).
Example: If f(x) = 5$x^4$, then find f'(x).
Solution: Let c = 5 and g(x) = $x^4$, then f'(x) = cg'(x) = 5(4$x^3$) = 20$x^3$.


4) Derivative of the sum of functions (sum rule).
The derivative of a function f(x) = g(x) + h(x) is given by f'(x) = g'(x) + h'(x)
Example: If f(x)= 5$x^4$ + 9, then find f'(x).
Solution: let g(x) = 5$x^4$ and h(x) = 9, then f'(x) = g'(x) + h'(x) = 20$x^3$ + 0 = 20$x^3$.


5) Derivative of the difference of functions.
The derivative of a function f(x)= g(x) - h(x)is given by f'(x) = g'(x) - h'(x).
Example: If f(x) = x 3 - x -2, then find f'(x).
Solution: Let g(x) = $x^5$ and h(x) = $x^{-2}$, then, f'(x) = g'(x) - h'(x) = 5$x^4$ - (-2 $x^{-3}$) = 5$x^4$ + 2$x^{-3}$.


6) Derivative of the product of two functions (product rule or Leibniz rule).
The derivative of a function f(x) = g(x) h(x) is given by f'(x) = g'(x) h(x) + g(x)h'(x).
It is also represented as (UV)’ = U’V + U V’.
Example: f(x) = ($x^3$ - 3x) (x - 3).
Solution: Let g(x) = ($x^3$ - 3x) and h(x) = (x - 3), then
f'(x) = g(x) h'(x) + h(x) g'(x) = (3$x^2$ - 3) (x-3) + ($x^3$ - 3x) (1)
=3 $x^3$ - 9 $x^2$ - 3x + 9 + $x^3$ - 3x = 4$x^3$ - 9$x^2$ - 3x + 9


7) Derivative of the quotient of two functions (quotient rule).
The derivative of a function f(x) = $\frac{g(x)}{h(x)}$ is given by f^'(x) = $\frac{g'(x)h(x)- g(x)h(x)}{(h(x))^2}$

Example: f(x) = $\frac{x – 5}{x + 7}$.

Solution: Let g(x) = (x - 5) and h(x) = (x + 7), then f'(x) = $\frac{g'(x)h(x)- g(x)h(x)}{(h(x))^2}$

= $\frac{((1)(x + 7) - (x - 5)(1))}{(x + 7)^2}$ = $\frac{(x + 7) - (x-5)}{(x+7)^2}$ = $\frac{x + 7 - x + 5}{(x+7)^2}$ = $\frac{12}{(x+7)^2}$.


8) Derivative of a logarithmic function.
The derivative of f(x) = $log_b$x is given by f'(x) = $\frac{1}{(x\ lnb)}$

If f(x) = ln x, then f'(x) = $\frac{1}{x}$.
Example: Find the derivative of f(x) = $log_5$x
Solution: Applying the formula mentioned above we get,

f'(x) = $\frac{1}{(x\ ln\ 5)}$.

Example: Find the derivative of f(x) = ln x + 3$x^2$
Solution: Let g(x) = ln x and h(x) = 3$x^2$, Using the sum rule, f'(x) = g'(x) + h'(x), we get

f'(x)= $\frac{1}{x}$ + 6x.


9) Derivative of an exponential function.
The derivative of an exponential function f(x)= $e^x$ with respect to x is f^'(x)= $e^x$
The derivative of an exponential function with base a, f(x) = $a^x$ with respect to x is f'(x) = $a^x$ ln a
Example: Find the derivative of f(x)= 7$e^x$.
Solution: The derivative of the function f(x)= 7$e^x$ is given by 7 f'($e^x$) = 7$e^x$.
Example: Find the derivative of f(x)= $7^x$.
Solution: The derivative of the function f(x)= $7^x$ is given by f'($7^x$) = $7^x$ ln 7.


10) Derivatives of Trigonometric functions.
The derivatives of the six trigonometric functions are tiven as below:
(sin x)’ = cos x. (cot x)’ = -$csc^2$ x.
(cos x)’ = -sin x. (sec x)’ = sec x tan x.
(tan x)’ = $sec^2$ x. (csc x)’ = -csc x cot x.


11) Derivatives of Inverse Trigonometric functions.

(arcsin x)' = $\frac{1}{\sqrt{1 - x^2}}$

(arccos x)' = $\frac{1}{\sqrt{1 - x^2}}$

(arctan x)' = $\frac{1}{\sqrt{1 + x^2}}$

(arcsec x)' = $\frac{1}{|x|\sqrt{x^2 - 1}}$

(arccsc x)' = $\frac{1}{|x|\sqrt{x^2 - 1}}$

(arccot x)' = $\frac{1}{1 + x^2}$


12) Derivatives of Hyperbolic functions.

(sinh x)' = cosh x = $\frac{e^x + e^{-x}}{2}$

(cosh x)' = sinh x = $\frac{e^x - e^{-x}}{2}$

(tanh x)' = sec$h^2$ x

(sech x)' = -tanh x sech x

(csch x)' = -coth x csch x

(coth x)' = -csc$h^2$ x

(arsinh x)' = $\frac{1}{\sqrt{x^2 + 1}}$

(arcosh x)' = $\frac{1}{\sqrt{x^2 - 1}}$

(artanh x)' = $\frac{1}{1 - x^2}$

(arsech x)' = -$\frac{1}{x \sqrt{1 - x^2}}$

(arcsch x)' = $\frac{1}{|x| \sqrt{1 + x^2}}$

(arcoth x)' = $\frac{1}{1 - x^2}$


13) Chain rule of differentiation.

Let f(x) = (g o h)(x) = g(h(x)) and u = g(x).
Let y = f(u); then the derivative of f with respect to x, f'(x) is given by:

f'(x) = ($\frac{dy}{du}$) ($\frac{du}{dx}$).

Example: Find the derivative f'(x), if f is given by f(x) = 5cos(2x - 3)
Solution: Let u = 2x - 3 and y = 5cos u, hence

$\frac{du}{dx}$ = 2 and $\frac{dy}{du}$ = -5 sin u

Using the chain rule, f'(x) = ($\frac{dy}{du}$)($\frac{du}{dx}$) = -5sin(u)(2) = -10sin(u)

Substituting back, u = 2x - 3 in sin(u), we get f'(x) = -10sin (2x - 3).