# What is differential calculus and how to find it?

In this post, we have discussed the definition, working, kinds, and rules of differential calculus along with examples. Now after reading the below basics of differential calculus, you can solve any problem of differentiation easily.

# What is differential calculus and how to find it?

In mathematics, differential calculus is one of the well-known branches of calculus that is widely used to find the differential of the function. the differential of the function can be found with respect to an independent variable.

This branch of calculus is also used to find the slope of the tangent line. The differential function can be calculated either by using rules of differentiation or the first principle method. In this article, we will learn the basics of differential calculus along with examples.

## Differential calculus

The differential calculus is frequently used in mathematical analysis concerned with the problem to determine the rate of change of function according to the independent variable. Differential calculus is generally used to find the differential of the functions and use them to evaluate the problems of non-constant rate of change.

The exponential, constant, trigonometric, polynomial, linear, logarithmic, or quadric functions are involved in differential calculus.

## Kinds of differential calculus

There are several kinds of derivatives in calculus. Let us discuss some of them.

### 1.   Explicit differentiation

Explicit differentiation is the main kind of differential calculus used to differentiate the single variable function with respect to the independent variable. This kind of differential calculus is denoted by f’(x) or d/dx. The independent variable could be x, y, z, u, v, w, t, etc.

### 2.   Partial differentiation

Partial differentiation is that kind of differential calculus that involves multivariable functions. It differentiates the function with respect to any variable available in the function. the partial derivative is denoted by ∂f(x, y, z)/∂x, ∂f(x, y, z)/∂y, and ∂f(x, y, z)/∂z.

### 3.   Implicit differentiation

Implicit differentiation is that type of differential calculus that deals with the equation and implicit functions. The implicit function is written as f(x, y) = g(x, y). The differentiation notation must be applied on both sides of the equation.

This kind of differentiation determines the derivative of the dependent variable with respect to the independent variable without considering it as a constant. It is denoted by sdy/dx or y’(x).

## Rules of differential calculus

Here are some commonly used rules of differential calculus.

1.     Sum rule: d/dw [g(w) + h(w)] = d/dw g(w) + d/dw h(w)

2.     Difference rule: d/dw [g(w) - h(w)] = d/dw g(w) - d/dw h(w)

3.     Constant rule: d/dw [c] = 0, where c is any constant

4.     Constant function rule: d/dw [c * f(w)] = cd/dw [f(w)], where c is any constant

5.     Power rule: d/dw [f(w)] n = n[f(w)] n-1

6.     Product rule: d/dw [g(w) * h(w)] = h(w) * [d/dw g(w)] + g(w) * [d/dw h(w)]

7.     Quotient rule: d/dw [g(w) / h(w)] = 1/(h(w))2 [h(w) * [d/dw g(w)] - g(w) * [d/dw h(w)]]

8.     Chain rule: dy/dw= [dy/du * du/dw]

## How to solve differential calculus problems?

Let us take a few examples to learn how to solve the problems of differential calculus.

Example 1

Find the differential of 4w3 + 6w2 + cos(w) – 3w + 9 with respect to w.

Solution

Step 1: First of all, apply the notation of differentiation to the given function.

d/dw [4w3 + 6w2 + cos(w) – 3w + 9]

Step 2: Now write the differential notation to each function separately by using the sum and difference rules of differential calculus.

d/dw [4w3 + 6w2 + cos(w) – 3w + 9] = d/dw [4w3] + d/dw [6w2] + d/dw [cos(w)] – d/dw [3w] + d/dw 

Step 3: Use the constant function rule of differential calculus and write the constant coefficients outside the differential notation.

d/dw [4w3 + 6w2 + cos(w) – 3w + 9] = 4 d/dw [w3] + 6 d/dw [w2] + d/dw [cos(w)] – 3 d/dw [w] + d/dw 

Step 4: Differentiate the above expression with respect to “w” with the help of power, trigonometric, and constant rules of differentiation.

d/dw [4w3 + 6w2 + cos(w) – 3w + 9] = 4 [3w3 – 1] + 6 [2w2 – 1] + [-sin(w) * dw/dw] – 3 [w1 – 1] + 

= 4 [3w2] + 6 [2w1] + [-sin(w) * dw/dw] – 3 [w0] + 

= 4 [3w2] + 6 [2w] + [-sin(w) * 1] – 3  + 

= 12w2 + 12w - sin(w) – 3

Use a differentiation calculator to find the differential of single variable function to avoid larger calculations.

Example 2: For implicit differentiation

Find the differential of 2wy – 3w2 + 2w – sin(y) = 2w2y + 4w – 24y + 2 w.r.t “w”.

Solution

Step 1: First of all, take the given equation and write the differentiation notation on it.

2wy – 3w2 + 2w – sin(y) = 2w2y + 4w – 24y + 2

d/dw [2wy – 3w2 + 2w – sin(y)] = d/dw [2w2y + 4w – 24y + 2]

Step 2: Now write the differential notation to each function separately by using the sum and difference rules of differential calculus.

d/dw [2wy] – d/dw [3w2] + d/dw [2w] – d/dw [sin(y)] = d/dw [2w2y] + d/dw [4w] – d/dw [24y] + d/dw 

Step 3: Use the product rule of differential calculus.

2y d/dw [w] + 2w d/dw [y] – d/dw [3w2] + d/dw [2w] – d/dw [sin(y)] = 2y d/dw [w2] + 2w2 d/dw [y] + d/dw [4w] – d/dw [24y] + d/dw 

Step 4: Differentiate the above expression with respect to “w” with the help of power, trigonometric, and constant rules of differentiation.

2y [w1 – 1] + 2w d/dw [y] – [3 * 2 w2 – 1] + [2 w1 – 1] – [cos(y) dy/dw] = 2y [2 w2 – 1] + 2w2 d/dw [y] + [4 w1 - 1] – d/dw [24y] + 

2y [w0] + 2w d/dw [y] – [3 * 2 w1] + [2 w0] – [cos(y) dy/dw] = 2y [2 w1] + 2w2 d/dw [y] + [4 w0] – d/dw [24y] + 

2y  + 2w d/dw [y] – [3 * 2 w] + [2 * 1] – [cos(y) dy/dw] = 2y [2 w] + 2w2 d/dw [y] + [4 * 1] – d/dw [24y] + 

2y + 2w d/dw [y] – [3 * 2 w] +  – [cos(y) dy/dw] = 2y [2 w] + 2w2 d/dw [y] +  – d/dw [24y] + 

2y + 2w dy/dw – [6 w] + 2 – cos(y) dy/dw = 4wy + 2w2 dy/dw + 4 – 24 dy/dw

Step 5: Take the dy/dw terms on the same side of the equation.

2w dy/dw – cos(y) dy/dw + 24 dy/dw - 2w2 dy/dw = 4wy + 6w - 2 + 4 – 2y

2w dy/dw – cos(y) dy/dw + 24 dy/dw - 2w2 dy/dw = 4wy + 6w + 2 – 2y

(2w – cos(y) + 24 - 2w2) dy/dw = 4wy + 6w + 2 – 2y

dy/dw = (4wy + 6w + 2 – 2y) / (2w – cos(y) + 24 - 2w2)

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