The derivatives are the measure of the function that how it changes as function’s input changes. Derivatives describe the behavior of output of a function with respect to any change in its input. Basically, a derivative indicates the nature of a function with the small change in its input value. We can say that, derivative is the ratio of infinitesimal change in output and infinitesimal change in input of a function. The process of finding derivative of a function is called differentiation.
For example: We know that the rate of change of velocity is known as acceleration. If we differentiate the function of velocity with respect to time, we get the function for acceleration. |

Let us consider a function $f(x)$ and $h$ be the infinitesimal change in its input. Then, the derivative of $f(x)$ is given by:

Derivative of a function f(x) with respect to x is denoted by either ${f}'$ or ${f}'(x)$ or by either $\frac{\mathrm{d} f(x)}{\mathrm{d} x}$ or $\frac{\mathrm{d} f}{\mathrm{d} x}$.

If differentiation is again imposed on the function f(x), then it is called

**second-order derivative**and is denoted by either ${f}''$ or ${f}''(x)$ or by either $\frac{\mathrm{d} {f}'}{\mathrm{d} x}$ or $\frac{\mathrm{d} f^{(1)}}{\mathrm{d} x}$.

Higher order derivatives can also be determined and are denoted accordingly.

Derivatives follow the following rules:

**• Constant Rule:**Let us assume a function $f$ as a function of $x$, then $\frac{\mathrm{d} (cf(x))}{\mathrm{d} x}$ = $c$ $\frac{\mathrm{d} f(x)}{\mathrm{d} x}$, where $c$ is a constant term.

**• Rule of Summation of Derivatives:**If we have two functions $f(x)$ and $g(x)$, then $\frac{\mathrm{d}}{\mathrm{d} x}\left \{ f(x)+g(x) \right \}$ = $\frac{\mathrm{d} f(x)}{\mathrm{d} x}$ + $\frac{\mathrm{d} g(x)}{\mathrm{d} x}$

**• Rule of Difference of Derivatives:**For two functions $f(x)$ and $g(x)$, $\frac{\mathrm{d}}{\mathrm{d} x}$ $\left \{ f(x)-g(x) \right \}$ = $\frac{\mathrm{d} f(x)}{\mathrm{d} x}$ - $\frac{\mathrm{d} g(x)}{\mathrm{d} x}$.

**• Chain Rule:**If $f$ is the function of $u$ and $u$ is the function of $x$, then derivative of $f$ with respect to $x$ is defined as: $\frac{\mathrm{d} f}{\mathrm{d} x}$ = $\frac{\mathrm{d} f}{\mathrm{d} u}$ $\times$ $\frac{\mathrm{d} u}{\mathrm{d} x}$.

**• Product Rule:**Consider $f$ and $g$ as the function of $x$, then according to the product rule: $\frac{\mathrm{d} }{\mathrm{d} x}$ $\left \{ f(x) \times g(x) \right \}$ = $f(x) \times$ $\frac{\mathrm{d} g(x)}{\mathrm{d} x}$ + $g(x) \times$ $\frac{\mathrm{d} f(x)}{\mathrm{d} x}$.

**• Quotient Rule:**$f$ and $g$ are two functions of $x$, then quotient rule says: $\frac{\mathrm{d} }{\mathrm{d} x}$ $\left \{ \frac{f(x)}{g(x)} \right \}$ = $\frac{g(x) \frac{df(x)}{dx} -f(x) \frac{dg(x)}{dx}}{ \{ g(x) \}^{2}}$.

We have studied so far in this page, about the derivatives of functions, which depend upon one variable. Sometimes, a function depends upon more than one variables. In this case, partial derivatives can be determined. Partial derivative of a function (which has two or more independent variables) is the derivative with respect to one variable in consideration with other variables as constant terms.

Let us consider a functions $f(x, y)$, that means $f$ is the function of $x$ and $y$ both. Then, partial derivative $f$ with respect to $x$ is the derivative of $f$ with respect to $x$, considering $y$ as a constant term. Similarly, partial derivative of f with respect to $y$ is the derivative of $f$ with respect to $y$, considering $x$ as constant. Partial derivative of a function $f$ is denoted by $\frac{\partial f}{\partial x}$.

The differentiation formulas and rules for finding partial derivative are same as that for finding normal derivative.

Let us consider an example.

### Solved Example

**Question:**Find the partial derivative of $f = x^{3} + 2xy + y^{3}$

**Solution:**

$\frac{\partial f}{\partial x}$ = $3x^{2} + 2y$ (considered y as constant)

and $\frac{\partial f}{\partial y}$ = $2x + 3y^{2}$ (considered x as constant).

and $\frac{\partial f}{\partial y}$ = $2x + 3y^{2}$ (considered x as constant).

- For a constant term c, $\frac{\mathrm{d}c }{\mathrm{d} x}$ = 0.
- For $a> 0$, $\frac{\mathrm{d} }{\mathrm{d} x}$$a^{x}$ = $a^{x}\ln a$.
- $\frac{\mathrm{d} }{\mathrm{d} x}$$x^{n} = nx^{n - 1}$.
- $\frac{\mathrm{d} }{\mathrm{d} x}$$\ln x$ = $\frac{1}{x}$.
- $\frac{\mathrm{d} }{\mathrm{d} x}$$e^{x} = e^{x}$.
- $\frac{\mathrm{d} }{\mathrm{d} x}$$\sin x = \cos x$.
- $\frac{\mathrm{d} }{\mathrm{d} x}$$\cos x = -\sin x$.
- $\frac{\mathrm{d} }{\mathrm{d} x}$$\tan x = \sec ^{2}x$.
- $\frac{\mathrm{d} }{\mathrm{d} x}$$\cot x = -\csc ^{2}x$.
- $\frac{\mathrm{d} }{\mathrm{d} x}$$\csc x = -\csc x \cot x$.
- $\frac{\mathrm{d} }{\mathrm{d} x}$$\sec x = \sec x \tan x$.
- $\frac{\mathrm{d} }{\mathrm{d} x}$$\sin^{-1}x$ = $\frac{1}{\sqrt{1 - x^{2}}}$.
- $\frac{\mathrm{d} }{\mathrm{d} x}$$\cos^{-1}x$ = $-$$\frac{1}{\sqrt{1 - x^{2}}}$.
- $\frac{\mathrm{d} }{\mathrm{d} x}$$\tan^{-1}x$ = $\frac{1}{1 + x^{2}}$.
- $\frac{\mathrm{d} }{\mathrm{d} x}$$\cot ^{-1}x = -$$\frac{1}{1 + x^{2}}$.
- $\frac{\mathrm{d} }{\mathrm{d} x}$$\csc ^{-1}x = -$$\frac{1}{x\sqrt{x^{2} - 1}}$.
- $\frac{\mathrm{d} }{\mathrm{d} x}$$\sec ^{-1}x$ = $\frac{1}{x\sqrt{x^{2} - 1}}$.

### Solved Examples

**Question 1:**If $y = 5x^{2} - x^{4} + \sqrt{x} + 5$, find $\frac{\mathrm{d} y}{\mathrm{d} x}$.

**Solution:**

$y = 5x^{2} - x^{4} + \sqrt{x} + 5$

$= 5x^{2} - x^{4} + x$$^{\frac{1}{2}}$$ + 5$

$\frac{\mathrm{d} y}{\mathrm{d} x} = 5 \frac{\mathrm{d} }{\mathrm{d} x}x^{2} - \frac{\mathrm{d} }{\mathrm{d} x}x^{4} + \frac{\mathrm{d} }{\mathrm{d} x}x^{\frac{1}{2}} + \frac{\mathrm{d} }{\mathrm{d} x}5$

= $5 \times 2x^{2 - 1} - 4x^{4 - 1} + $$\frac{1}{2}x^{\frac{1}{2} - 1}$$ + 0$

= $10x - 4x^3 + $$\frac{1}{2}x^{-\frac{1}{2}}$

= $10x - 4x^3 + $$\frac{1}{2\sqrt{x}}$

$= 5x^{2} - x^{4} + x$$^{\frac{1}{2}}$$ + 5$

$\frac{\mathrm{d} y}{\mathrm{d} x} = 5 \frac{\mathrm{d} }{\mathrm{d} x}x^{2} - \frac{\mathrm{d} }{\mathrm{d} x}x^{4} + \frac{\mathrm{d} }{\mathrm{d} x}x^{\frac{1}{2}} + \frac{\mathrm{d} }{\mathrm{d} x}5$

= $5 \times 2x^{2 - 1} - 4x^{4 - 1} + $$\frac{1}{2}x^{\frac{1}{2} - 1}$$ + 0$

= $10x - 4x^3 + $$\frac{1}{2}x^{-\frac{1}{2}}$

= $10x - 4x^3 + $$\frac{1}{2\sqrt{x}}$

**Question 2:**Let $y = e^{x} + \cos x$, find $\frac{\mathrm{d} y}{\mathrm{d} x}$.

**Solution:**

$y = e^{x} + \cos x$

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}e^{x} + \frac{\mathrm{d} }{\mathrm{d} x}\cos x$

= $e^{x} + (-\sin x)$

= $e^{x} - \sin x$

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}e^{x} + \frac{\mathrm{d} }{\mathrm{d} x}\cos x$

= $e^{x} + (-\sin x)$

= $e^{x} - \sin x$

Derivatives are widely used in commerce for finding marginal profit, marginal cost, marginal revenue etc. Derivatives are used in the physics quite often. Many formulas and practical applications are based upon the knowledge of derivatives. Derivatives are used in electronics and other engineering branches. Few examples based on derivatives are shown below:

### Solved Examples

**Question 1:**Find $\frac{\mathrm{d} y}{\mathrm{d} x}$, if $y = e^{x} \sin x + 7 \tan x$.

**Solution:**

$y = e^{x} \sin x + 7 \tan x$

Using product rule in first term:

$\frac{\mathrm{d} y}{\mathrm{d} x} = e^{x}\frac{\mathrm{d} }{\mathrm{d} x} \sin x + \sin x \frac{\mathrm{d} }{\mathrm{d} x}e^{x} + 7 \frac{\mathrm{d} }{\mathrm{d} x} \tan x$

$= e^{x} \cos x + \sin x e^{x} + 7 \sec ^{2}x$

$= e^{x} (\cos x + \sin x) + 7 \sec ^{2}x$

Using product rule in first term:

$\frac{\mathrm{d} y}{\mathrm{d} x} = e^{x}\frac{\mathrm{d} }{\mathrm{d} x} \sin x + \sin x \frac{\mathrm{d} }{\mathrm{d} x}e^{x} + 7 \frac{\mathrm{d} }{\mathrm{d} x} \tan x$

$= e^{x} \cos x + \sin x e^{x} + 7 \sec ^{2}x$

$= e^{x} (\cos x + \sin x) + 7 \sec ^{2}x$

**Question 2:**Given that $y = \sqrt{\cos x}$. Differentiate y with respect to x.

**Solution:**

$y = \sqrt{\cos x}$

$y = \cos$$^{\frac{1}{2}}$$x$

Using chain rule:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{2}\cos ^{\frac{1}{2}-1}x\ \times \frac{\mathrm{d} }{\mathrm{d} x}\cos x$

= $\frac{1}{2\sqrt{\cos x}}$$(-\sin x)$

= -$\frac{\sin x}{2\sqrt{\cos x}}$

$y = \cos$$^{\frac{1}{2}}$$x$

Using chain rule:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{2}\cos ^{\frac{1}{2}-1}x\ \times \frac{\mathrm{d} }{\mathrm{d} x}\cos x$

= $\frac{1}{2\sqrt{\cos x}}$$(-\sin x)$

= -$\frac{\sin x}{2\sqrt{\cos x}}$