There are various branches of mathematics in daily life and the educational system. Calculus is one of the main branches of mathematics that is frequently used to deal with the properties of differentiation and integration.

The derivative and integral are the types of calculus that are used to calculate the slope of the tangent line and area under the curve respectively. In this lesson, we will discuss one main type of calculus which is derivative.

## What is a derivative

One of the main branches of calculus that deals with the instantaneous rate of change of a function with respect to the independent variable. It is generally used to calculate the slope of the tangent line.

Derivative calculus can also be defined with the help of limit calculus by a first principle method. The process of finding the rate of change of function is known as differentiation. It is also known as differential calculus.

Differentiation is one of the main way*s to calculate the rate of change in any quantity and the other terms in differentiation. The linear, algebraic, quadric, trigonometric, hyperbolic, inverse, and constant functions can be differentiated with respect to the independent variable.*

*There are further types of dif*ferentiation such as explicit differentiation, implicit differentiation, partial differentiation, and directional differentiation that are frequently used to calculate the differential of single, double, multivariable functions, and direction of the function respectively.

It is denoted by f’(w) or d/dw. There is a general equation for finding the derivative of functions with the help of the first principle method. Such as

**d/dw**** [f(w****)****]**** = lim**_{h}_{→}_{0}** f(w + h) – f(w****) / h**

## Rules of differentiation

**Let us ****discuss**** the ****general and basic rules of differentiation that are widely used to calculate derivative problems. ve calculus?**

The derivative is o

Rules name | Rule |

Trigonometry Rule | d/dt [sin(t)] = cos(t) d/dt [cos(t)] = -sin(t) d/dt [tan(t)] = sec ^{2}(t) |

Power Rule | d/dt [f(t)]^{n} = n [f(t)]^{n-1} * d/dt [f(t)] |

Quotient Rule | d/dt [f(t) / g(t)] = 1/[g(t)]^{2} [g(t) d/dt [f(t)] – f(t) d/dt [g(t)]] |

Product Rule | d/dt [f(t) * g(t)] = g(t) d/dt [f(t)] + t(t) d/dt [g(t)] |

Constant Rule | d/dt [L] = 0, where L is any constant |

Difference Rule | d/dt [f(t) – g(t)] = d/dt [f(t)] – d/dt [g(t)] |

Sum Rule | d/dt [f(t) + g(t)] = d/dt [f(t)] + d/dt [g(t)] |

Exponential Rule | d/dt [e^{t}] = e^{t} |

Constant function Rule | d/dt [L * f(t)] = L d/dt [f(t)] |

## How to differentiate the functions?

The problems of differential calculus can be solved easily either by using the first principle method or rules of differentiation. Let us take some examples of calculating the problems of differentiation by the first principle and rules of differentiation.

**Example 1: By ****the ****first principle method**

Calculate the differential of the given function with respect to “t”.

f(t) = t^{2} + 4t + 8

**Solution**

**Step 1:** First of all, write the general expression of the first principle method.

d/dt [f(t)] = lim_{h}_{→}_{0} f(t + h) – f(t) / h

**Step 2:** Now the given function and the addition of the** ****h** function to the above formula.

f(t) = t^{2} + 4t + 8

f(t + h) = (t + h)^{2} + 4(t + h) + 8 = t^{2} + 2ht + h^{2} + 4t + 4h + 8

f(t + h) = t^{2} + 2ht + h^{2} + 4t + 4h + 8

**substitution**

d/dt [t^{2} + 4t + 8] = lim_{h}_{→}_{0} [t^{2} + 2ht + h^{2} + 4t + 4h + 8] – [t^{2} + 4t + 8] / h

d/dt [t^{2} + 4t + 8] = lim_{h}_{→}_{0} [t^{2} + 2ht + h^{2} + 4t + 4h + 8 – t^{2} – 4t – 8] / h

d/dt [t^{2} + 4t + 8] = lim_{h}_{→}_{0} [2ht + h^{2} + 4h] / h

d/dt [t^{2} + 4t + 8] = lim_{h}_{→}_{0} [2t + h + 4]

**Step 3: **Now apply the limit value h = 0 to the above expression.

d/dt [t^{2} + 4t + 8] = lim_{h}_{→}_{0} [2t] + lim_{h}_{→}_{0} [h] + lim_{h}_{→}_{0} [4]

d/dt [t^{2} + 4t + 8] = [2t] + [0] + [4]

d/dt [t^{2} + 4t + 8] = 2t + 4

**Example**** 2: By using the laws**

Differentiate the given function with respect to “u”.

f(u) = 4u^{3} + 3u^{2} – 12u * 15u^{2} + 20

**Solution**

**Step ****1:** Take the given function and independent variable after that apply the notation of differentiation to it.

f(u) = 4u^{3} + 3u^{2} – 12u * 15u^{2} + 20

corresponding variable = u

d/du f(u) = d/du [4u^{3} + 3u^{2} – 12u * 15u^{2} + 20]

**Step 2****:** Now use the sum and difference law of differentiation to apply the differential notation to each term of the function separately.

d/du [4u^{3} + 3u^{2} – 12u * 15u^{2} + 20] = d/du [4u^{3}] + d/du [3u^{2}] – d/du [12u * 15u^{2}] + d/du [20]

**Step 3:** Now apply the product rule of differentiation.

d/du [4u^{3} + 3u^{2} – 12u * 15u^{2} + 20] = d/du [4u^{3}] + d/du [3u^{2}] – [15u^{2 }d/du [12u] + 12u d/du [15u^{2}]] + d/du [20]

d/du [4u^{3} + 3u^{2} – 12u * 15u^{2} + 20] = d/du [4u^{3}] + d/du [3u^{2}] – 15u^{2 }d/du [12u] – 12u d/du [15u^{2}] + d/du [20]

**Step ****3:** Now take the constant coefficients outside the differential notation.

d/du [4u^{3} + 3u^{2} – 12u * 15u^{2} + 20] = 4d/du [u^{3}] + 3d/du [u^{2}] – 15u^{2 }d/du [12u] – 12u d/du [15u^{2}] + d/du [20]

**Step-****4:** Now differentiate the above expression with the help of the power rule.

= 4 [3 u^{3-1}] + 3 [2 u^{2-1}] – 15u^{2} [12 u^{1-1}] – 12u [30 u^{2-1}] + [0]

= 4 [3 u^{2}] + 3 [2 u^{1}] – 15u^{2} [12 u^{0}] – 12u [30 u^{1}] + [0]

= 4 [3 u^{2}] + 3 [2 u] – 15u^{2} [12 (1)] – 12u [30 u] + [0]

= 12u^{2} + 6u – 180u^{2} – 360u^{2}

= 6u – 528u^{2}

= 6u (1 – 88u)

To ease up the larger calculations, a derivative calculator can be used. This calculator will give you the step-by-step solution to the given problem in a single click.

**Example**** 3****: By using the rule****s**

Calculate the differential of the given function with respect to “w”.

f(w) = 8w^{2} / 12w – 21w – 5w^{2} + 10w

**Solution**

**Step ****1:** Take the given function and independent variable after that apply the notation of differentiation to it.

f(w) = 8w^{2} / 12w – 21w – 5w^{2} + 10w

corresponding variable = w

d/dw f(w) = d/dw [8w^{2} / 12w – 21w – 5w^{2} + 10w]

**Step 2****:** Now use the sum and difference law of differentiation to apply the differential notation to each term of the function separately.

d/dw [8w^{2} / 12w – 21w – 5w^{2} + 10w] = d/dw [8w^{2} / 12w] – d/dw [21w] – d/dw [5w^{2}] + d/dw [10w]

**Step 3:** Now apply the quotient rule of differentiation.

d/dw [8w^{2} / 12w – 21w – 5w^{2} + 10w] = 1/(12w)^{2} [12w d/dw [8w^{2}] – 8w^{2} d/dw [12w]] – d/dw [21w] – d/dw [5w^{2}] + d/dw [10w]

**Step ****4****:** Now take the constant coefficients outside the differential notation.

d/dw [8w^{2} / 12w – 21w – 5w^{2} + 10w] = 1/(12w)^{2} [12w d/dw [8w^{2}] – 8w^{2} d/dw [12w]] – 21d/dw [w] – 5d/dw [w^{2}] + 10d/dw [w]

**Step-****5****:** Now differentiate the above expression with the help of the power rule.

= 1/(12w)^{2} [12w [16w^{2-1}] – 8w^{2} [12w^{1-1}]] – 21 [w^{1-1}] – 5 [2 w^{2-1}] + 10 [w^{1-1}]

= 1/(12w)^{2} [12w [16w^{1}] – 8w^{2} [12w^{0}]] – 21 [w^{0}] – 5 [2 w^{1}] + 10 [w^{0}]

= 1/(12w)^{2} [12w [16w] – 8w^{2} [12(1)]] – 21 [1] – 5 [2 w] + 10 [1]

= 1/(12w)^{2} [12w [16w] – 8w^{2} [12]] – 21 – 5 [2w] + 10

= 1/(144w^{2}) [192w^{2} – 96w^{2}] – 21 – 10w + 10

= 192w^{2}/(144w^{2}) – 96w^{2}/(144w^{2}) – 21 – 10w + 10

= 96/72 – 48/72 – 21 – 10w + 10

= 8/6 – 4/6 – 11 – 10w

= 4/3 – 2/3 – 11 – 10w

= 2/3 – 11 – 10w

= -31/3 – 10w

## Wrap up

Now we have covered all the basic intent of the derivative calculus in this lesson. Now you can calculate any problem of differentiation either by the first principle or by using rules of differentiation just by learning the above post.