fbpx
MATHEMATICS TUTORIAL

How To Calculate Differential Calculus Problems?

How to solve calculus
How to solve calculus
Written by Nkedugist

How to solve calculus in mathematics with proper format and easy guidelines check here. In mathematics, there are two well-known branches of differential calculus, one is integral calculus and the other is differential calculus. The integral calculus is used to find the area under the curve.

While the differential calculus is used to find the slope of the tangent line. the differential calculus uses various rules of derivative and the lmit by the first principle to solve its

In this post, we will discuss how to calculate the problems of differential calculus What is differential calculus?

In mathematics, the differential is the general branch of calculus used to differentiate the single, double, or multivariable functions according to the respective variable.

In simple words, the rate of change of single, double, or multivariable functions with respect to the independent variable is known as the differential calculus.

The functions of differentiation can be exponential, constant, linear, logarithmic, trigonometric, polynomial, or quadric.
The first principle method of differentiation uses the below expression

Limh→0 f(x + h) – f(x) / h
Rule of differentiation
Constant rule: d/dh [k] = 0
Power rule: d/dh [f(h)]n = n * [f(h)]n – 1 d/dh [f(h)] Constant function rule: d/dh [k * f(h)] = k * d/dh [f(h)] Sum rule: d/dh [f(h) + g(h)] = d/dh [f(h)] + d/dh [g(h)] Difference rule: d/dh [f(h) – g(h)] = d/dh [f(h)] – d/dh [g(h)] Exponential rule: d/dh [eh] = eh
Product rule: d/dh [f(h) * g(h)] = g(h) d/dh [f(h)] + f(h) d/dh [g(h)] Quotient rule: d/dh [f(h) / g(h)] = 1/[g(h)]2 [g(h) d/dh [f(h)] – f(h) d/dh [g(h)]] Logarithmic rule: d/dh [ln (h)] = 1/h

Kinds of differentiation

There are various kinds of differentiation in calculus. Let us briefly discuss some of them.

1. Implicit differentiation

It is the type of differentiation in calculus used to solve equation-related functions. The implicit differentiation uses the f(h, y) = g(h, y) form to write an implicit function of two variables.

This kind of differentiation is usually used to find the differential of a dependent variable with respect to the independent variable.

In simple words, the implicit derivative is used to find the differential of “y” concerning the independent variable “h” such as dy/dh.

2. Explicit differentiation

Explicit differentiation is usually used to find the differential of variable function with respect to its independent variable. This is the most commonly used type of differentiation. The main derivative problems are solved by using this kind of differentiation.
If the independent variable is “h” then it is denoted by d/dh.

3. Partial differentiation

This kind of differentiation is commonly used to solve the problem of two, three, or more variable functions with respect to one of the independent variables.

The multivariable function can be written as f(h, u, y) and is denoted by:
∂f(h, u, y)/∂h, ∂f(h, u, y)/∂u, and ∂f(h, u, y)/∂y

How to solve the problems of differential calculus?

The rules of differentiation are very helpful in solving the derivative of functions. Below are a few solved examples of differentiation to learn how to calculate it.

Example 1
Evaluate the explicit differentiation of 3h3 + 2h2 – 7h4 + 4sin(h) + 2h + 19 with respect to “h”

Solution

Step 1: First of all, take the given function f(h) and apply the differential notation to it.
f(h) = 3h3 + 2h2 – 7h4 + 4sin(h) + 2h + 19
d/dh [f(h)] = d/dh [3h3 + 2h2 – 7h4 + 4sin(h) + 2h + 19]

Step 2: Use the sum and difference rules of differential calculus and write the differential notation with each function separately.

d/dh [3h3 + 2h2 – 7h4 + 4sin(h) + 2h + 19] = d/dh [3h3] + d/dh [2h2] – d/dh [7h4] + d/dh [4sin(h)] + d/dh [2h] + d/dh [19]

Step 3: Now write all the constant coefficients outside the notation of differential with the help of the constant function rule.

= 3 d/dh [h3] + 2 d/dh [h2] – 7 d/dh [h4] + 4 d/dh [sin(h)] + 2 d/dh [h] + d/dh [19]

Step 4: Now differentiate the above expression by using the constant, trigonometric, & power rules of differentiation.

= 3 [3h3-1] + 2 [2h2-1] – 7 [4h4-1] + 4 [cos(h)] + 2 [h1-1] + [0] = 3 [3h2] + 2 [2h1] – 7 [4h3] + 4 [cos(h)] + 2 [h0] + [0] = 3 [3h2] + 2 [2h] – 7 [4h3] + 4 [cos(h)] + 2 [1] + [0] = 3 [3h2] + 2 [2h] – 7 [4h3] + 4 [cos(h)] + 2
= 9h2 + 4h – 28h3 + 4cos(h) + 2

To ease up the calculations, use a derivative calculator to get the result with steps in a fraction of seconds.

Example 2
Evaluate 3hcos(h) + 6h2 – 27h3 + 15y2 + 9hy = 19y – 3hy with respect to “y”.

Solution

Step 1: Write the given function and apply the notation of the derivative.
f(h, y) = g(h, y)
3hcos(h) + 6h2 – 27h3 + 15y2 + 9hy = 19y – 3hy

d/dy [3hcos(h) + 6h2 – 27h3 + 15y2 + 9hy] = d/dh [19y – 3hy]

Step 2: Apply d/dx to each function separately by using the sum & difference rules of differential and write the constant coefficients outside the differential notation.

d/dy [3hcos(h)] + d/dy [6h2] – d/dy [27h3] + d/dy [15y2] + d/dy [9hy] = d/dh [19y] – d/dy [3hy]

Step 3: Now write all the constant coefficients outside the notation of differential with the help of the constant function rule.

3 d/dy [h * cos(h)] + 6 d/dy [h2] – 27 d/dy [h3] + 15 d/dy [y2] + 9 d/dy [h * y] = 19 d/dh [y] – 3 d/dy [h * y]

Step 4: Now differentiate the above expression by using the constant, trigonometric, product, & power rules of differentiation.

3 cos(h) d/dy [h] + 3h d/dh [cos(h)] + 6 [2 h2-1] – 27 [3h3-1] + 15 [2y2-1 d/dh (y)] + 9 [y * d/dh (h) + h d/dh (y)] = 19 d/dh [y] – 3 [y * d/dh (h) + h d/dh (y)]

3 cos(h) [h1-1] + 3h [-sin(h)] + 6 [2 h1] – 27 [3h2] + 15 [2y1 d/dh (y)] + 9 [y * (h1-1) + h d/dh (y)] = 19 d/dh [y] – 3 [y * (h1-1) + h d/dh (y)]

3 cos(h) [h0] + 3h [-sin(h)] + 6 [2 h] – 27 [3h2] + 15 [2y d/dh (y)] + 9 [y * (h0) + h d/dh (y)] = 19 d/dh [y] – 3 [y * (h0) + h d/dh (y)]

3 cos(h) [1] + 3h [-sin(h)] + 6 [2 h] – 27 [3h2] + 15 [2y d/dh (y)] + 9 [y * (1) + h d/dh (y)] = 19 d/dh [y] – 3 [y * (1) + h d/dh (y)]

3 cos(h) + 3h [-sin(h)] + 6 [2 h] – 27 [3h2] + 15 [2y d/dh (y)] + 9 [y + h d/dh (y)] = 19 d/dh [y] – 3 [y + h d/dh (y)]

3cos(h) – 3hsin(h) + 12h – 81h2 + 30y dy/dh + 9y + 9h dy/dh = 19 dy/dh – 3y + 3h dy/dh

Step 5: Now separate the dy/dh terms.

30y dy/dh + 9h dy/dh – 19 dy/dh – 3h dy/dh = – 3y – 3cos(h) + 3hsin(h) – 12h + 81h2 – 9y

(30y + 9h – 19 – 3h) dy/dh = – 3y – 3cos(h) + 3hsin(h) – 12h + 81h2 – 9y

dy/dh = (–3y – 3cos(h) + 3hsin(h) – 12h + 81h2 – 9y) / (30y + 9h – 19 – 3h)

Summary

this post, we have discussed the type, rules, and examples to learn how to calculate the problems of differential calculus. Now after reading this article, you can grab all the basics of differential calculus.

Facebooktwitterredditpinterestlinkedinmail

About the author

Nkedugist

I created Nkedugist.com.ng to serve and assist all Aspirants who are seeking school information or want to apply for admission into Nigerian Tertiary Institutions such as Universities, Polytechnics or Colleges of Education with information regarding the various education system.

Leave a Comment

You cannot copy content of this page! Contact the Publisher

Skip to toolbar