Differentiation in Math: What is it? Explained with Calculations
Calculus is one of the main branches of mathematics along with geometry, trigonometry, algebra, and number system. It is a specific method for solving various kinds of problems to find the area under the curve and the slope of the tangent line.
There are two well-known branches of calculus, one is differentiation and the other is integration. Both types of calculus play a vital role in the calculation of complex math problems.
What is differentiation?
In calculus, differentiation is the process of finding the differential of a function. The differential is the instantaneous rate of change of the dependent variable of the function with respect to the independent variable of the function.
It is frequently used to measure the per unit change in the independent variable of the function. For example, if a function h = g(u) is given then the rate of change of “h” with respect to the dependent variable “u” can be written as:
d/du [h] or d/du [f(u)]
Differentiation can also be done with the help of limits such as:
d/du [f(u)] = Limh→0 [f(u + h) – f(u) / h]
Notations of differentiation
There are different notations used to denote the differential of the function.
Name | Symbol |
Leibniz’s Notation | d/du [f(u)] |
Euler’s Notation | D(h) or D[f(u)] |
Lagrange’s Notation | F’(u) |
Rules of differentiation
Rules of differentiation are very essential for calculating the rate of change of a function.
Rules Name | Expression |
Rule of Power | d/du [h(u)]n = n h(u)n-1 * d/du [h(u)] |
Rule of Constant | d/du [C] = 0 |
Rule of Sum | d/du [h(u) + g(u)] = d/du [h(u)] + d/du [g(u)] |
Rule of Product | d/du [h(u) * g(u)] = g(u) d/du [h(u)] + h(u) d/du [g(u)] |
Rule of Difference | d/du [h(u) + g(u)] = d/du [h(u)] + d/du [g(u)] |
Rule of Quotient | d/du [h(u) / g(u)] = 1/[g(u)]2 [g(u) d/du [h(u)] – h(u) d/du [g(u)]] |
Rules of Trigonometry | d/du [cos(u)] = -sin(u) d/du [sin(u)] = cos(u) d/du [tan(u)] = sec2(u) d/du [cot(u)] = -cosec2(u) d/du [sec(u)] = Sec(u) tan(u) d/du [cosec(u)] = -cosec(u) cot(u) d/du [cos2(u)] = -2cos(u) sin(u) d/du [sin2(u)] = 2sin(u) cos(u) |
How to solve the differentiation problem
The problems of differentiation can be solved easily with the help of rules of differentiation. There is another way to solve the problems of differentiation is by getting help from online resources.
Example 1
Find the differential of f(u) = 6u3 + 4u3 – 3sin2(u) + 12u – 6u4 + tan(x) with respect to “u”.
Solution
Step 1: First of all, apply the notation of differentiation to the given expression.
D [f(u)] = D [6u3 + 4u3 – 3sin2(u) + 12u – 6u4 + tan(x]
Step 2: Use the rules of differentiation to apply the notation of differentiation with each term of the above expression.
D [6u3 + 4u3 – 3sin2(u) + 12u – 6u4 + tan(x] = D [6u3] + D [4u3] – D [3sin2(u)] + D [12u] – D [6u4] + D [tan(x]
Step 3: Now take out the coefficients of the above expression.
D [6u3 + 4u3 – 3sin2(u) + 12u – 6u4 + tan(x] = 6D [u3] + 4D [u3] – 3D [sin2(u)] + 12D [u] – 6D [u4] + D [tan(x]
Step 4: Now apply the rules of power, constant, and trigonometry to evaluate the differentiation of the given expression.
D [6u3 + 4u3 – 3sin2(u) + 12u – 6u4 + tan(x] = 6 [3u3-1] + 4 [3u3-1] – 3 [2sin2-1(u) D (sin(u))] + 12 [u1-1] – 6 [4u4-1] + [0]
D [6u3 + 4u3 – 3sin2(u) + 12u – 6u4 + tan(x] = 6 [3u2] + 4 [3u2] – 3 [2sin1(u) (cos(u))] + 12 [u0] – 6 [4u3] + [0]
D [6u3 + 4u3 – 3sin2(u) + 12u – 6u4 + tan(x] = 6 [3u2] + 4 [3u2] – 3 [2sin(u) (cos(u))] + 12 [1] – 6 [4u3]
D [6u3 + 4u3 – 3sin2(u) + 12u – 6u4 + tan(x] = 18u2 + 12u2 – 6sin(u) * cos(u) + 12 – 24u3
D [6u3 + 4u3 – 3sin2(u) + 12u – 6u4 + tan(x] = 30u2 – 6sin(u) * cos(u) + 12 – 24u3
D [6u3 + 4u3 – 3sin2(u) + 12u – 6u4 + tan(x] = 30u2 – 6sin(u) * cos(u) + 12 – 24u3
Alternatively
Solved through derivative calculator by MeraCalculator.
Example 1
Find the differential of f(v) = 12v4 – 5v2 + 4sin(v) – 2cos(v) + 12v – 12v4 + 15 with respect to “v”.
Solution
Step 1: First of all, apply the notation of differentiation to the given expression.
D [f(v)] = D [12v4 – 5v2 + 4sin(v) – 2cos(v) + 12v – 12v4 + 15]
Step 2: Use the rules of differentiation to apply the notation of differentiation with each term of the above expression.
D [12v4 – 5v2 + 4sin(v) – 2cos(v) + 12v – 12v4 + 15] = D [12v4] – D [5v2] + D [4sin(v)] – D [2cos(v)] + D [12v] – D [12v4] + D [15]
Step 3: Now take out the coefficients of the above expression.
D [12v4 – 5v2 + 4sin(v) – 2cos(v) + 12v – 12v4 + 15] = 12D [v4] – 5D [v2] + 4D [sin(v)] – 2D [cos(v)] + 12D [v] – 12D [v4] + D [15]
Step 4: Now apply the rules of power, constant, and trigonometry to evaluate the differentiation of the given expression.
D [12v4 – 5v2 + 4sin(v) – 2cos(v) + 12v – 12v4 + 15] = 12 [4 v4-1] – 5 [2 v2-1] + 4 [cos(v)] – 2 [-sin(v)] + 12 [v1-1] – 12 [4 v4-1] + [0]
D [12v4 – 5v2 + 4sin(v) – 2cos(v) + 12v – 12v4 + 15] = 12 [4 v3] – 5 [2 v1] + 4 [cos(v)] – 2 [-sin(v)] + 12 [v0] – 12 [4 v3] + [0]
D [12v4 – 5v2 + 4sin(v) – 2cos(v) + 12v – 12v4 + 15] = 12 [4 v3] – 5 [2 v1] + 4 [cos(v)] – 2 [-sin(v)] + 12 [1] – 12 [4 v3]
D [12v4 – 5v2 + 4sin(v) – 2cos(v) + 12v – 12v4 + 15] = 48v3 – 10v + 4cos(v) + 2sin(v) + 12 – 48 v3
D [12v4 – 5v2 + 4sin(v) – 2cos(v) + 12v – 12v4 + 15] = 48v3 – 10v + 4cos(v) + 2sin(v) + 12 – 48 v3
D [12v4 – 5v2 + 4sin(v) – 2cos(v) + 12v – 12v4 + 15] = – 10v + 4cos(v) + 2sin(v) + 12
Alternatively
Solved through derivative calculator by limitcalculator.online
FAQs
What is differentiation?
It's the process of calculating the differential of a function in calculus. The differential is the instantaneous rate of change between the dependent and intended variables of the function.
How many types of differentiation?
There are different types of differentiation such as explicit differentiation, implicit differentiation, partial differentiation, and directional differentiation. These types are used to find the derivative of a single variable function, a derivative of the dependent variable w.r.t independent variable, a derivative of a multivariable function, and the direction of the derivative respectively.
What is the relation between differentiation and integration?
Differentiation and integration are two well-known branches of calculus. The integral is the reverse process of differentiation. It means integral finds the original function of a differentiated function.
What is the formula of the first principle method?
The formula for the first principle is:
D [f(u)] = Limh→0 [f(u + h) – f(u) / h]
Final Words
Differentiation is a well-known technique in calculus that is essential to find the slope of the tangent line. There are different notations of differential calculus. The rules of differentiation are very helpful for evaluating the problems of differentiation.
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