Derivative in Calculus & Method of Calculation

Derivative in calculus: In calculus, the derivative is frequently used to find the differential of functions and the slope of the tangent line. The function of the differential could be a single variable, double variable (implicit function), or multivariable.

Rules of differentiation or the first principle method are the methods to find the differential of the functions. Limit is also a kind of calculus used to define continuity, integration, and differentiation. In this post, we will learn all the basics such as definitions, types, and rules of derivatives along with examples.

What is derivative in calculus?

In calculus, the derivative is the rate of change of a function with respect to its independent variable. The derivative is frequently used in mathematical analysis to concern the problems of differential calculus.

The derivative is the reverse process of integral calculus. It means the differentiation reverses the terms that an integral does.  The derivative could be calculated for the constant, exponential, logarithmic, trigonometric, single variable (linear), multivariable (polynomial), etc.

 Types of derivative

There are several types of derivatives used to find the derivative of different kinds of functions. Let us discuss the types of differential along with the solved examples.

1.   Explicit derivative

In calculus, explicit differentiation is the major type of derivative used to solve the one variable functions with respect to the respective variables. This type of differentiation is a well-known method of determining the rate of change of single variable functions.

It is denoted by d/dx, where x is the independent variable. The independent variable could be x, y, z, u, v, w, t, etc. The general expression of this type of differential calculus is:

d/dx [f(x)] = F(x)

Let us take an example of explicit differentiation.

Example: For explicit differentiation

Calculate 5u² + 9u³ – 18u⁴ + 4cos(u) + 2u + 19 with respect to “u”.

Solution

Step I: First of all, write the given differential function and apply the notation of differentiation on it.

h(u) = 5u² + 9u³ – 18u⁴ + 4cos(u) + 2u + 19

d/du [h(u)] = d/du [5u² + 9u³ – 18u⁴ + 4cos(u) + 2u + 19]

Step II: Now apply the notation of differentiation to each function separately by using the sum, difference, and the trigonometric rules of differentiation.

d/du [5u² + 9u³ – 18u⁴ + 4cos(u) + 2u + 19] = d/du [5u²] + d/du [9u³] – d/du [18u⁴] + d/du [4cos(u)] + d/du [2u] + d/du [19]

step III: Now apply the constant function rule of differential calculus.

d/du [5u² + 9u³ – 18u⁴ + 4cos(u) + 2u + 19] = 5d/du [u²] + 9d/du [u³] – 18d/du [u⁴] + 4d/du [cos(u)] + 2d/du [u] + d/du [19]

Step IV: Now find the derivative of the above expression by with respect to “u”.

d/du [5u² + 9u³ – 18u⁴ + 4cos(u) + 2u + 19] = 5[2u^2-1]+9[3u^3-1]– 18[4u^4-1] + 4[-sin(u)] + 2[u^1-1]+ [0]

= 5 [2u¹] + 9 [3u²] – 18 [4u³] + 4 [-sin(u)] + 2 [u⁰] + 0

= 5 [2u] + 9 [3u²] – 18 [4u³] + 4 [-sin(u)] + 2 [1] + 0

= 5 [2u] + 9 [3u²] – 18 [4u³] – 4 [sin(u)] + 2

= 10u + 27u² – 72u³ – 4sin(u) + 2

The above problem of the explicit differentiation can also be solved by using a derivative calculator.

2.   Implicit differentiation

In calculus, to solve the implicit function or dependent function with respect to the independent function implicit differentiation is used. This type of derivative is used to find the dy/dx of the function.

It is denoted by dy/du or y’(u). It means you have to find the derivative of y with respect to x such as the derivative of y² is 2ydy/dx. The general expression of implicit differentiation is

d/du f(u, y) = d/du g(u, y)

Let us take an example of explicit differentiation.

Example: For implicit differentiation

Calculate 6u + ycos(u) + 6u² – 27y³ + 15u² = y + 12u + 20 with respect to “u”.

Solution

Step I: First of all, write the given implicit function and apply the notation of differentiation on it.

d/du [6u + ycos(u) + 6u² – 27y³ + 15u²] = d/du [y + 12u + 20y² + 2u]

Step II: Now apply the notation of differentiation to each function separately by using the sum, difference, product, and the trigonometric rules of differentiation.

d/du [6u] + d/du [ycos(u)] + d/du [6u²] – d/du [27y³] + d/du [15u²] = d/du [y] + d/du [12u] + d/du [20y²] + d/du [2u]

d/du [6u] + cos(u) d/du [y] + y d/du [cos(u)] + d/du [6u²] – d/du [27y³] + d/du [15u²] = d/du [y] + d/du [12u] + d/du [20y²] + d/du [2u]

step III: Now apply the constant function rule of differential calculus.

6d/du [u] + cos(u) d/du [y] + y d/du [cos(u)] + 6d/du [u²] – 27d/du [y³] + 15d/du [u²] = d/du [y] + 12d/du [u] + 20d/du [y²] + 2d/du [u]

Step IV: Now find the derivative of the above expression by with respect to “u”.

6 [u^1-1] + cos(u) [dy/du] + y [-sin(u)] + 6 [2u^2-1] – 27 [3y^3-1 dy/du] + 15 [2u^2-1] = dy/du + 12 [u^1-1] + 20 [2y^2-1 dy/du] + 2 [u^1-1]

6 [u⁰] + cos(u) [dy/du] + y [-sin(u)] + 6 [2u¹] – 27 [3y² dy/du] + 15 [2u¹] = dy/du + 12 [u⁰] + 20 [2y¹ dy/du] + 2 [u⁰]

6 [1] + cos(u) [dy/du] + y [-sin(u)] + 6 [2u] – 27 [3y² dy/du] + 15 [2u] = dy/du + 12 [1] + 20 [2y dy/du] + 2 [1]

6 + cos(u) [dy/du] – y [sin(u)] + 6 [2u] – 27 [3y² dy/du] + 15 [2u] = dy/du + 12 + 20 [2y dy/du] + 2

6 + cos(u) dy/du – y * sin(u) + 12u – 81y² dy/du + 30u = dy/du + 12 + 40y dy/du + 2

Step V: Now take the dy/du terms to the one side of the equation and take it common.

cos(u) dy/du– 81y² dy/du – dy/du – 40y dy/du = 12 + 2 – 6 – 12u – 30u + y * sin(u)

[cos(u) – 81y² – 1 – 40y] dy/du = 8 – 42u + y * sin(u)

dy/du = [8 – 42u + y * sin(u)] / [cos(u) – 81y² – 1 – 40y]

y’(u) = [8 – 42u + y * sin(u)] / [cos(u) – 81y² – 1 – 40y]

Summary

In this post, we have covered the two main types of derivatives along with general expressions and solved examples. Now after reading the above post, you can easily solve any problem of explicit and implicit differentiation.

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