# Introduction to Derivatives and how to differentiate equations?

In calculus, differentiation is a basic concept that is used to find the derivative of the function. Calculus is the branch of mathematics that is used to find the change in functions. Before calculus, all the terms were calculated statistically.

Differentiation is one of the main topics of calculus. There are many applications of differentiation in engineering such as the voltage across a capacitor and many other engineering topics like how quickly the quantities change.

In this post, we will discuss differentiation and how to solve it with examples.

#### What is differentiation?

A technique or a method that is used to analyze the changes in the given function is known as differentiation. In simple words, the instantaneous rate of change of the functions with respect to independent variables is called differentiation. It is also helpful to measure how quickly a function is changing at any point.

It is also known as differential in calculus because its work is to calculate the derivative of the given function in terms of the independent variables. It is mainly used to find the slope of the tangent line.

It is denoted by $\frac{d}{dx}$ to find the rate of change of the function. Differentiation can also be calculated by using the limits known as the first principle method.

$\frac{d}{dx} f(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

#### How to apply differentiation on different equations?

For the calculation of derivative problems, you must be familiar with the rules and formulas of derivatives. The problems of derivatives are dependent on the rules and formulas. Let us take some examples to know how to solve them?

Example 1: By first principal method

Find the derivative of $2x$ by using the first principle method?

Solution

Step 1: Take the given function and write it equal to $f(x)$.

$f(x) = 2x$

Step 2: Add h in the above function.

$f (x + h) = 2(x + h)$

Step 3: Now take the general formula.

$\frac{d}{dx} f(x) = \lim_{h \to 0}\frac{ f (x + h) - f(x)}{h}$

Step 4:Put the values in the above formula.

$\frac{d}{dx} f(x) = \lim_{h \to 0}\frac{ 2(x + h) - 2x}{h}$

$\frac{d}{dx} f(x) = \lim_{h \to 0}\frac{ 2x + 2h - 2x }{ h}$

$\frac{d}{dx} f(x) = \lim_{h \to 0}\frac{ 2h}{h}$

$\frac{d}{dx} f(x) = \lim_{h \to 0} 2$

Step 5: Apply limit.

$\frac{d}{dx} f(x) = 2$

Because the limit of a constant function remains unchanged.

Step 6: Result

$\frac{d}{dx} 2x = 2$

Example 2

Find the derivative of $2x^2 + 3x + 3$ by using the first principle method.

Solution

Step 1: Take the given function and write it equal to $f(x)$.

$f(x) = 2x^2 + 3x + 3$

Step 2: Add $h$ in the above function in each $x$ term.

$f (x + h) = 2(x + h)^2 + 3(x + h) + 3$

Step 3: Now take the general formula.

$\frac{d}{dx} f(x) = \lim_{h \to 0}\frac{ f (x + h) - f(x) }{ h}$

Step 4: Put the values in the above formula.

$\frac{d}{dx} f(x) = \lim_{h \to 0}\frac{ 2(x + h)^2 + 3(x + h) + 3 - (2x^2 + 3x + 3) }{ h}$

$\frac{d}{dx} f(x) = \lim_{h \to 0}\frac{ 2(x^2 + 2hx + h^2) + 3(x + h) + 3 - (2x^2 + 3x + 3) }{ h}$

$\frac{d}{dx} f(x) = \lim_{h \to 0}\frac{ 2x^2 + 4hx + 2h^2 + 3x + 3h + 3 - 2x^2 - 3x - 3 }{ h}$

$\frac{d}{dx} f(x) = \lim_{h \to 0}\frac{ 4hx + 2h^2 + 3h }{ h}$

$\frac{d}{dx} f(x) = \lim_{h \to 0}\frac{ h (4x + 2h + 3) }{ h}$

$\frac{d}{dx} f(x) = \lim_{h \to 0}(4x + 2h + 3)$

Step 5: Apply limit.

$\frac{d}{dx} f(x) = \lim_{h \to 0} (4x + 2h + 3)$

$\frac{d}{dx} f(x) = \lim_{h \to 0} 4x + 2(0) + 3$

$\frac{d}{dx} f(x) = \lim_{h \to 0} 4x + 3$

Step 6: Write the result with the question.

$\frac{d}{dx} (2x^2 + 3x + 3) = 4x + 3$

Example 3: By using rules

Find the derivative of $2x^4 + 3x^3 - 3x \cdot 4x^2 + 2$ by using rules of differentiation.

Solution

Step 1: Take the given function and write it equal to $f(x)$.

$f(x) = 2x^4 + 3x^3 - 3x \cdot 4x^2 + 2$

Step 2: Apply $\frac{d}{dx}$ on both sides.

$\frac{d}{dx} f(x) = \frac{d}{dx} (2x^4 + 3x^3 - 3x \cdot 4x^2 + 2)$

Step 3: Now apply the rules.

$\frac{d}{dx} f(x) = \frac{d}{dx} (2x^4) + \frac{d}{dx} (3x^3) - \frac{d}{dx} (3x \cdot 4x^2) + \frac{d}{dx} (2)$

Step 4: Now solve the differentiation.

$\frac{d}{dx} f(x) =\frac{d}{dx} (2x^4) + \frac{d}{dx} (3x^3) - \frac{d}{dx} (3x \cdot 4x^2) + \frac{d}{dx} (2)$

$\frac{d}{dx} f(x) = (2 \cdot 4) x^{4-1} + (3 \cdot 3) x^{3-1} - (3x \frac{d}{dx} (4x^2) + 4x^2 \frac{d}{dx} (3x)) + 0$

$\frac{d}{dx} f(x) = 8x^3 + 9x^2 - (3x (4 \cdot 2) x^{2-1} + 4x^2 (3x^{1-1}))$

$\frac{d}{dx} f(x) = 8x^3 + 9x^2 - (3x (8x) + 4x^2 (3))$

$\frac{d}{dx} f(x) = 8x^3 + 9x^2 - 24x^2 - 12x^2$

$\frac{d}{dx} f(x) = 8x^3 + 9x^2 - 36x^2$

$\frac{d}{dx} f(x) = 8x^3 - 27x^2$

You can also use a derivative calculator to get the output of your problem in a step-by-step way.

https://www.meracalculator.com/math/derivative.php

Example 4

Find the derivative of $5x^5 + 9x^2 - \frac{13x}{12x^3} + 121$ by using rules of differentiation.

Solution

Step 1: Take the given function and write it equal to $f(x)$.

$f(x) = 5x^5 + 9x^2 - \frac{13x}{12x^3} + 121$

Step 2: Apply $\frac{d}{dx}$ on both sides.

$\frac{d}{dx} f(x) = \frac{d}{dx} (5x^5 + 9x^2 - \frac{13x}{12x^3} + 121)$

Step 3: Now apply the rules.

$\frac{d}{dx} f(x) = \frac{d}{dx}(5x^5) + \frac{d}{dx} (9x^2) - \frac{d}{dx}\left(\frac{13x}{12x^3}\right) + \frac{d}{dx} (121)$

Step 4: Now solve the differentiation.

$\frac{d}{dx} f(x) = \frac{d}{dx} (5x^5) + \frac{d}{dx} (9x^2) - \frac{d}{dx} \left(\frac{13x }{12x^3}\right) + \frac{d}{dx} (121)$

$\frac{d}{dx} f(x) = (5 \cdot 5) x^{5-1} + (9 \cdot 2) x^{2-1} - \frac{1}{(12x^3)^2} \left[12x^3 \frac{d}{dx} (13x) - 13x \frac{d}{dx} (12x^3)\right] + 0$

$\frac{d}{dx} f(x) = 25x^4 + 18x - \frac{1}{144x^6} \left[12x^3 (13x^{1-1}) - 13x (12 \cdot 3) x^{3-1}\right]$

$\frac{d}{dx} f(x) = 25x^4 + 18x - \frac{1}{144x^6} (12x^3 (13) - 13x (36) x^2)$

$\frac{d}{dx} f(x) = 25x^4 + 18x - \frac{1}{144x^6} (156x^3 - 468x^3)$

$\frac{d}{dx} f(x) = 25x^4 + 18x - \frac{1}{144x^6}(- 312x^3)$

$\frac{d}{dx} f(x) = 25x^4 + 18x + \frac{312x^3}{144x^6}$

$\frac{d}{dx} f(x) = 25x^4 + 18x + \frac{312x^3}{144x^6}$

$\frac{d}{dx} f(x) = 25x^4 + 18x + \frac{13}{6x^3}$

#### Summary

Now you can easily solve derivative problems by using any method mentioned above. By following the above examples, you can grab the concept of solving the derivative problems in short period of time.

Derivative solver with steps

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