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).
The derivative of a product of two functions U and V is also represented as (UV)' = U'V + UV'.Example: f(x) = $x^3$ Sinx.

Solution: Let U = $x^3$ and V = Sinx. Then the differentiated product is given by
($x^3$ Sinx)' = ($x^3$)' Sinx + ($x^3$)(Sinx)' = 3$x^2$ Sinx + $x^3$Cosx.
In differential calculus, logarithmic differentiation or differentiation by taking logarithms is a method used to differentiate functions by applying the logarithmic derivative of a function f.
[ln(f)]' = $\frac{f'}{f}$ $\rightarrow$ f' = f .[ln(f)]' Logarithmic differentiation is often performed in cases where it is easier to differentiate the logarithm of a function rather than the function itself. This technique relies on the chain rule as well as properties of logarithms to transform products into sums and divisions into subtractions, and can also be applied to functions raised to the power of variables or other functions.

Solved Example

Question: Use the method of logarithmic differentiation to find the derivative, y', if y is given by y = $x^{sin\ 2x}$.

Solution:
 
Using the method of logarithmic differentiation, we first take the natural logarithm on both sides of
                   y = $x^{sin2x}$.

ln y = ln [$x^{sin2x}$] .

Using  logarithm properties we can rewrite as,

ln y = sin2x ln x .

Differentiating both sides of the above equation with respect to x,
(using the chain rule on the left side of equation and the product rule formula for differentiation on the right side of equation).

$\frac{1}{y}$ y' = (2cos2x) ln x + sin2x ($\frac{1}{x}$).

Multiplying both sides by y we get,

y' = [2cos2x ln x + ($\frac{1}{x}$)sin2x] y .

y' = [2cos2x ln x + ($\frac{1}{x}$)sin2 x ] $x^{sin x}$.