Differentiation and Integration are both quite crucial concepts in calculus which are typically used to learn the change. Calculus is not only restricted to mathematics but has a huge array of applications in various domains of science as well as the economy. Also, we may be able to spot calculus in establishing an analysis in finance as well as in the stock market. In this chapter, we will study some differentiation and integration formula with examples besides the interesting concept!

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Find below some of the basic formulas of differentiation and integration.

The mathematical concept of Differentiation facilitates us to find rates of change. For example, it enables us to detect the rate of change of velocity with respect to time (called the acceleration). In addition, It also enables us to find the rate of change of x with reference to y, which graphically at y against x is the gradient of the curve. There are a set of simple rules which can be used in order to differentiate a number of functions easily.

If y = some function of x, then the derivative of y (in respect to x) is written as dy/dx.

For Example: Find out the gradient of the curve y = 2x4 at the point (4, 56)?

Using the formula, dy/dx = 8 x2

When x = 4, dy/dx = 8 × 7 = 56.

Integration is a mathematical technique to find a function g(x) the derivative of which, Dg(x), is equivalent to a provided function f(x). This is denoted by the integral sign “∫,” or ∫f(x), generally termed the indefinite integral of the function. The sign dx denotes a displacement of an infinitesimal along x; therefore ∫f(x) dx becomes the aggregate sum of the product of f(x) and dx. The definite integral is written as: - “∫ab

With ‘a’ and ‘b’ referred to the limits of integration, is equivalent (=) to g(b) − g(a), where Dg(x) = f(x).

Integration is a mathematical approach of adding slices to find the whole. Applying the theory of Integration, we are able to easily find areas, volumes, median points of many useful figures. That said, it is still simplest to start with finding the area under the curve of a function like one below:

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Differentiation under the integral sign is an algebraic operation in calculus that is performed in order to assess certain integrals. Under a reasonably loose situation on the function being integrated, this operation enables us to swap the order of integration and differentiation. In its simplified version, called the Leibniz integral rule, differentiation under the integral sign models the ensuing equation legitimate under the formula on:

\[\frac{d}{dt} \int_{a}^{b} f(x, t) dx = \int_{a}^{b} \partial_{t} f(x, t) dx\]

Many integrals that would otherwise be impracticable or need a substantially more complex approach can be solved by this method.

Differentiation under the integral sign rule can be used to assess the certain unusual definite integrals such as given below:

\[I’(b) = \frac{d}{db} \int_{0}^{1} \frac{x^{b} - 1}{ln(x)} dx\]

\[\frac{d}{db} \int_{0}^{1} \frac{x^{b} - 1}{ln(x)} dx = \int_{0}^{1} \frac{\partial}{\partial b} \frac{x^{b} - 1}{ln(x)}\]

\[I’(b) = \int_{0}^{1} \frac{ln(x) x^{b}}{ln(x)} dx\]

Example1: Differentiate the given functions

\[Y = \sqrt[3]{x^{2}} (2x - x^{2})\]

Solution1:

In the given function we would not be able to just differentiate the 1st term, and then differentiate the 2nd term and multiply the two together. This would just NOT work in the case given. However, it is still possible to perform the derivative of this function. All we are supposed to do is convert the radical into fractional exponents and multiply this across the parenthesis.

\[Y= x^{\frac{2}{3}} (2x - x^{2}) = 2 x^{\frac{5}{3}} - x^{\frac{8}{3}}\]

Here, we can differentiate the function

\[Y^{1} = \frac{10}{3} x^{\frac{2}{3}} - \frac{8}{3} x^{\frac{5}{3}}\]

Example2: Find out if the given equation

\[f(x) = 2x^{3} + \frac{300}{x^{-3}} + 4\] rising, reducing or not changing at x= −2?

Solution2:

We are already aware that the rate of change of a function is provided by the derivative of the function. Thus all we are supposed to do is to rewrite the function. Rewriting the function is required in order to deal with the 2nd term and then take the derivative.

\[F(x) = 2x^{3} + \frac{300}{x^{-3}} + 4\] → f^{1} (x) = 6x^{2} - 900^{x^{-4}} = 6x^{2} - \frac{900}{x^{4}\]

Remember that had rewritten the last term in the derivative back as a fraction.

This is not something we’ve performed up to this step and is only being done here to help with the assessment in the next step. Note that it’s often simpler to do the assessment with positive exponents.

So, upon assessing the derivative we now obtain,

\[f^{1} (-2) = 6 (4) - \frac{900}{16} = - \frac{129}{4} = - 32.25\]

Thus, at x = −2, the derivative turns out to be negative and, hence, the function is decreasing at x = −2

FAQ (Frequently Asked Questions)

1. What is Meant by Integration by Parts?

Ans. Integration by parts is a unique mathematical technique of integration which is most commonly used when two functions are multiplied together but is also quite useful in other ways. It is actually a technique of using incorporating the product rule in reverse. Below is the rule of Integration by Parts:

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Where, u is the function, denoted by u(x)

v is the function, denoted by v(x)

u' is the derivative of the function, denoted by u(x)

Moreover,

The formula for Integration by Parts is as given:-

∫u dv = u. v −∫v du

2. How do we Find the Derivative?

Ans. You might find it really difficult, but it isn't. The trick is to first simplify the given expression: perform the division (divide each term on the numerator by 3x^{½}. Use the law of indices in what equation we get. Now, differentiate term by term. Note that there are different ways of writing the derivative. They are although all typically the same. In other words:

(1) If y = x^{3},

dy/dx = 2x

This implies that if y = x^{3} the derivative of y, in respect to x is 3x

(2) d/dx (x^{3}) = 2x

This implies that the derivative of x^{3} in respect to x is 2x.