Differential calculus is one of the two parts of calculus. The other part is integral calculus.
The main objects of differential calculus are to find derivatives of given functions which is known as differentiation process and related parts like differential equations along with their applications.

The derivative of a function is the measure which showcases the rate at which the given function changes with changes made in the independent variable. The derivative of a function with respect to 'x' is also the slope of the graph of that particular function. "$\frac{dy}{dx}$"or "y" are the notations that are used for representing the derivative of a function y in x with respect to x. since power is 1, so we call it first derivative. As we move on with differentiation again and again the powers increase and so does we call them with their respective powers.

Formulas

We have been given some common formulas that are used for finding derivatives of simple functions:

1) $\frac{d}{dy}$ $(y^n) = n y^{n -1}$

2) $\frac{d}{dy}$ $(m y^n)$ = $m$ * $\frac{d}{dy}$ $(y^n)$ where ‘m’ is a constant

3) $\frac{d}{dy}$ $a = 0$ where 'a' is a constant

4) $\frac{d}{dy}$ $(e^y)$ = $e^y$

5) $\frac{d}{dy}$ $(a^y)$ = $a^y$ . ln a

6) $\frac{d}{dy}$ $(ln y)$ = $\frac{1}{y}$ y > 0

7) $\frac{d}{dy}$ $(log_a (y))$ = $\frac{1}{(y\ ln\  a)}$

8) $\frac{d}{dy}$ $sin y = cos y$

9) $\frac{d}{dy}$ $cos y$ = - $sin y$

10) $\frac{d}{dy}$ $tan y$ = sec$^2$ y

11) $\frac{d}{dy}$ $sec y$ = sec y tan y

12) $\frac{d}{dy}$ $cot y$ = -$csc^2 y$

13) $\frac{d}{dy}$ $csc y$ = -$csc y\  cot y$

14) $\frac{d}{dx}$ $(sin^{-1} y)$ = $\frac{1}{\sqrt {1 - y^2}}$

15) $\frac{d}{dx}$ $(tan^{-1} y)$ = $\frac{1}{(1 + y^2)}$

16) $\frac{d}{dx}$ $(sec^{-1} y)$ = $\frac{1}{|y| \sqrt{y^2 - 1}}$

17) $\frac{d}{dy}$ $f (g (y))$ = $\frac{d}{dy}$ (f (g (y))) $\frac{d}{dy}$ (g (y))

18) $\frac{d}{dy}$ $(f (y) + g (y))$ = $\frac{d}{dy}$ (f (y)) + $\frac{d}{dy}$ (g (y))

19) $\frac{d}{dy}$ (f(y) - g(y)) = $\frac{d}{dy}$ (f(y)) - $\frac{d}{dy}$ (g(y))

For other compound functions we can make use of the product rule and the quotient rule.

Examples

Let us see some examples on differential calculus.

Example 1: Determine the second derivative of the following

$x^3 + cos (x^3) - 7x + 6$

Solution:

Let $y = x^3 + cos (x^3) – 7x + 6$

Differentiating both sides with respect to ‘x’ we get,

$\frac{dy}{dx}$ = $\frac{d}{dx}$ $(x^3 + cos (x^3) - 7x + 6)$

= $\frac{d}{dx}$ $(x^3)$ + $\frac{d}{dx}$ $(cos (x^3))$ - $\frac{d}{dx}$ (7x) + $\frac{d}{dx}$ 6

= 3 $x^2$ - $sin (x^3)$ $\frac{d}{dx}$ $(x^3)$ – 7 + 0

= 3 $x^2$ - 3 $x^2$ $sin (x^3)$ - 7

Differentiating with respect to ‘x’, we get,

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

= $\frac{d}{dx}$ $(3 x^2)$ - $\frac{d}{dx}$ $(3 x^2 sin (x^3))$ - $\frac{d}{dx}$

= 3 . 2 . x - [3 $x^2$ $\frac{d}{dx}$ $(sin (x^3) + sin (x^3)$ $\frac{d}{dx}$ $(3 x^2)$] - 0

= 6 x - 3 $x^2$ cos $(x^3)$ $(3 x^2)$ - 6 $x sin (x^3)$

= 6x -  9 $x^4 cos (x^3)$ - 6$x sin (x^3)$.

Example 2: Find y'', if y = 10 + $x^4$ + $cos (x^3)$ - tan(2x)

Solution:

Since y = 10 + $x^4$ + $cos (x^3)$ - tan(2x)

Differentiating both sides with respect to ‘x’ we get,

$\frac{dy}{dx}$ = $\frac{d}{dx}$ (10 + $x^4$ + $cos (x^3)$ - tan(2x))

$\rightarrow$    $\frac{dy}{dx}$ = $\frac{d}{dx}$ (10) + $\frac{d}{dx}$ ($x^4$) + $\frac{d}{dx}$ ($cos (x^3)$) - $\frac{d}{dx}$ (tan(2x))

$\rightarrow$    $\frac{dy}{dx}$ = 0 + 4x$^3$ - 3x$^2$sin (x$^3$) - 2sec$^2$ (2x)

$\rightarrow$    $\frac{dy}{dx}$ = 4x$^3$ - 3x$^2$sin (x$^3$) - 2sec$^2$ (2x)