In mathematics, several branches are used to solve various kinds of problems. These branches are algebra, geometry, calculus, set theory, etc. Calculus is one of the main branches of mathematics that deal with the study of differentiation and integration.
The subtypes of calculus are very essential in finding the solution to complex problems. In this post, we are going to describe differentiation along with calculations.
What is differentiation?
In calculus, the process of finding the derivative of a function w.r.t its independent variable and considering the dependent variables as a constant is said to be the differentiation. In geometry, it is widely used to find the slope of the tangent line.
It is a wide concept that is very helpful in finding complex calculus problems. There are two ways in differentiation for solving the derivative of the function, one is by the first principle method and the other is by using the rules of differentiation.
Formula of differentiation
The formula for finding the differential of the function by using limits is:
d/dx (f(x)) = limh→0 [f(x + h) – f(h)] / h
Laws of differentiation
There are various laws of differentiation that are helpful in solving complex calculus problems. Here are a few laws of differentiation.
1. Constant Law
The constant law of differentiation states that the derivative of any constant function gives zero as an output. Hence the derivative of any constant coefficient and dependent variables always give zero.
d/dx [k] = 0
where k is any constant.
2. Sum law
In differentiation, sum law is frequently used in complex calculus problems. according to this law, the notation of differential must be applied to each function separately. Whenever 2, 3, or more functions are given with a plus sign among them then the notation is applied separately.
d/dx [f(x) + p(x) + q(x)] = d/dx [f(x)] + d/dx [p(x)] + d/dx [q(x)]
3. Difference law
In differentiation, difference law is frequently used in complex calculus problems. According to this law, the notation of differential must be applied to each function separately. Whenever 2, 3, or more functions are given with a minus sign among them then the notation is applied separately.
d/dx [f(x) – p(x) – q(x)] = d/dx [f(x)] – d/dx [p(x)] – d/dx [q(x)]
4. Product law
The product law is widely used in differentiation to calculate the derivative of two or more functions with a multiply sign among them. According to this law, the differential is applied to the first function while the second function remains the same.
After that, the differential function is applied to the second function while the function remains the same with a plus sign between them.
d/dx [f(x) * p(x)] = p(x) d/dx [f(x)] + f(x) d/dx [p(x)]
5. Quotient law
The quotient law is widely used in differentiation to calculate the derivative of two or more functions with a division sign among them. According to this law, the differential is applied to the first function while the second function remains the same.
After that, the differential function is applied to the second function while the function remains the same with a minus sign between them. And divide it by the square of the second function.
d/dx [f(x) / p(x)] = 1/[p(x)]2 [p(x) d/dx [f(x)] – f(x) d/dx [p(x)]]
6. Power Law
In differentiation, the power law is used with the functions present with exponents. According to this law, the exponent must multiply with the function and the power must be reduced by 1.
d/dx [f(x)]m = m[f(x)]m-1 d/dx [f(x)]
How do calculate differentiation problems?
The problems of differentiation can be solved easily with the help of its laws. Let us take a few examples to learn how to calculate differentiation problems.
Example 1
Evaluate the differential of the given function w.r.t “x”.
p(x) = 2x4 + 5x6 – 12x2 + 3x + 12t
Solution
Step-1: First of all, apply the notation of differentiation to the given function.
d/dt p(x) = d/dt [2x4 + 5x6 – 12x2 + 3x + 12t]
Step-2: Use the sum and difference laws of differentiation to apply the notation separately to each function.
d/dt [2x4 + 5x6 – 12x2 + 3x + 12t] = d/dt [2x4] + d/dt [5x6] – d/dt [12x2] + d/dt [3x] + d/dt [12t]
Step-3: Now use the constant function law to take the constant coefficients outside the notation.
d/dt [2x4 + 5x6 – 12x2 + 3x + 12t] = 2d/dt [x4] + 5d/dt [x6] –`12d/dt [x2] + 3d/dt [x] + d/dt [12t]
Step-4: Use the power rule to find the differential of the above function.
= 2 [4 x4-1] + 5 [6 x6-1] –`12 [2 x2-1] + 3 [x1-1] + [0]
= 2 [4 x3] + 5 [6 x5] –`12 [2 x1] + 3 [x0] + [0]
= 2 [4 x3] + 5 [6 x5] –`12 [2 x] + 3 [1] + [0]
= 8 [x3] + 30 [x5] –`24 [x] + 3 [1]
= 8x3 + 30x5 –`24x + 3
A differentiation calculator can also be used to solve the above problem to ease up the calculations.
Example 2
Evaluate the differential of the given function w.r.t “u”.
p(u) = 5u2 + 12cos(u) – 15u12 + 5sin(u) + 2u3
Solution
Step-1: First of all, apply the notation of differentiation to the given function.
p(u) = 5u2 + 12cos(u) – 15u12 + 5sin(u) + 2u3
d/du p(u) = d/du [5u2 + 12cos(u) – 15u12 + 5sin(u) + 2u3]
Step-2: Use the sum and difference laws of differentiation to apply the notation separately to each function.
d/du [5u2 + 12cos(u) – 15u12 + 5sin(u) + 2u3] = d/du [5u2] + d/du [12cos(u)] – d/du [15u12] + d/du [5sin(u)] + d/du [2u3]
Step-3: Now use the constant function law to take the constant coefficients outside the notation.
d/du [5u2 + 12cos(u) – 15u12 + 5sin(u) + 2u3] = 5d/du [u2] + 12d/du [cos(u)] – 15d/du [u12] + 5d/du [sin(u)] + 2d/du [u3]
Step-4: Use the power rule to find the differential of the above function.
= 5 [2 u2-1] + 12 [-sin(u)] – 15 [12 u12-1] + 5 [cos(u)] + 2 [3 u3-1]
= 5 [2 u1] + 12 [-sin(u)] – 15 [12 u11] + 5 [cos(u)] + 2 [3 u2]
= 5 [2 u] + 12 [-sin(u)] – 15 [12 u11] + 5 [cos(u)] + 2 [3 u2]
= 10 [u] – 12 [sin(u)] – 180 [u11] + 5 [cos(u)] + 6 [u2]
= 10u – 12sin(u) – 180u11 + 5cos(u) + 6u2
Wrap up
In this post, we have discussed all the basics of differentiation and methods for solving it either by using an online calculator or manually. Now you can complete assignments and prepare for your exams without any difficulty.