Application of Derivatives & Indefinite Integration

In this article, we shall learn about the properties of derivatives and their application to find the maxima and minima. After that, we begin with integral calculus – indefinite integrals, properties, and methods of finding the integral of a function.

Also, check: Differentiation of a Function.

Check the latest NEET Syllabus for the 2023/2024 exams.

Table of Contents:

Properties of Derivatives

Following are the properties of derivatives that are very helpful while determining the derivative of a given function.

Derivative of a Constant

Let a be any real constant number and f be a function x then,

\(\begin{array}{l}\frac{d}{dx}a=0\;\;\;whereas\;\;\;\frac{d}{dx}af(x)=a\frac{d}{dx}f(x)\end{array} \)

For example, derivative of 5sin x is

d/dx (5sin x) = 5 d/dx(sin x) = 5 cos x.

Derivative of Sum or Difference of a Function

Suppose we have to find the derivative of the sum or difference of two functions of x, f(x) and g(x), then,

\(\begin{array}{l}\frac{d}{dx}(f(x)\pm g(x))=\frac{d}{dx}f(x)\pm \frac{d}{dx}g(x)\end{array} \)

For example, let us find the derivative of 4x2 + cosec x

d/dx [4x2 + cosec x] = d/dx(4x2) + d/dx(cosec x) = 8x – cosec x cot x.

Product Rule

The derivative of the product of two functions, f(x) and g(x,) is determined by

\(\begin{array}{l}\frac{d}{dx}[f(x).g(x)]=g(x)\frac{d}{dx}f(x)+ f(x)\frac{d}{dx}g(x)\end{array} \)

For example, let us find the derivative of x3cos x

d/dx(x3 cos x) = cos x d/dx(x3) + x3 d/dx(cos x)

= cos x. 3x2 + x3 (– sin x) = 3x2 cos x – x3 sin x

Quotient Rule

If two functions of x, f(x) and g(x) are in are in a division form, then

\(\begin{array}{l}\frac{d}{dx}\left [ \frac{f(x)}{g(x)} \right ]=\frac{\frac{df(x)}{dx}g(x)-\frac{dg(x)}{dx}f(x)}{(g(x))^{2}}\end{array} \)

For example, the derivative of (sin x)/(x2 + 1) is calculated as

\(\begin{array}{l}\frac{d}{dx}\frac{sin\;x}{(x^{2}+1)}=\frac{(x^{2}+1)\frac{d\;sin\;x}{dx}-sin\;x\frac{d(x^{2}+1)}{dx}}{(x^{2}+1)^{2}}\end{array} \)
\(\begin{array}{l}=\frac{(x^{2}+1)cos\;x-2x\;sin\;x}{(x^{2}+1)^{2}}\end{array} \)

Double Differentiation

Double differentiation is also known as the second-order derivative of a function. Double differentiation means differentiating the derivative of a function again. Let y = f(x) be a real-valued function, then the second order derivative of f is given by

\(\begin{array}{l}\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}\left [ \frac{dy}{dx} \right ]\end{array} \)

Double differentiation of f is also written as f’’(x).

The graphical interpretation of the second-order derivative of a function is that if the first-order derivative represents the slope of the function at a point, then the second-order derivative represents the rate of change of the slope of the graph.

Let us take an example, let f(x) = sin x, then the second-order derivative of f can be calculated as

\(\begin{array}{l}\frac{d^{2}}{dx^{2}}sin\;x=\frac{d}{dx}\left [ \frac{d}{dx} sin\;x\right ]=\frac{d}{dx}cos\;x=-sin\;x\end{array} \)

In this way, differentiating successively, we can find other higher-order derivatives of a given function.

Increasing and Decreasing Functions

One of the applications of derivatives is to find the nature of a function, whether it is increasing or decreasing.

A function f(x) is said to be an increasing function, if for any x1 < x2 we have f(x1) ≤ f(x2) which means for increasing values of x, the value of f(x) also increases.

In contrast,a function f(x) is said to be a decreasing function, if for any x1 < x2 we have f(x1) ≥ f(x2) which means for increasing values of x, the value of f(x) decreases.

For example, f(x) = ex is an increasing function.

Increasing function

And f(x) = e-x is a decreasing function.

Decreasing function

Now we may come across functions that increase for a certain interval and decrease for another. For example, f(x) = sin x,

Graph of sin x

We observe that for x ∈ [0, 𝜋/2], the function increases, and after that, it decreases. It also has some points where the tangent to the function has zero slope, that is, f’(x) = 0. These points are called critical points.

For increasing parts of the function, f’(x) > 0 and for decreasing parts of the function, f’(x) < 0.

Thus, to sum up:

If f’(x) > 0,

The function has an increasing tangent slope, that is, the function is increasing.

If f’(x) = 0,

The tangent to the function at x is parallel to the x-axis, and x is called a critical point of f.

If f’(x) < 0,

The function has a decreasing tangent slope, that is, the function is decreasing.

Increasing and decreasing function

Maxima and Minima

Maxima and minima are points in the domain of a function where the function attains maximum and minimum values, respectively.

Maxima and minima

Second Order Derivative Test

We shall use derivatives to determine local maxima and minima for a given function.

For a function f, if x = a is a maxima, then as the value of x increases after a, the slope of the tangent to the graph decreases. If x = a is a minima, then as the value of x increases after a, the slope of the tangent to the graph increases.

Thus, the second order derivative test is given as:

  • Let x = a be a critical point of f such that f’(x) = 0 at x = a, then if f’’(x) < 0, x = a is a point of local maxima.
  • Let x = a be a critical point of f such that f’(x) = 0 at x = a, then if f’’(x) > 0, x = a is a point of local minima.
  • If f’’(a) = 0, then the test fails.

Steps to find maxima and minima of a given function:

  • Let f(x) be the given function, find the first order derivative of f and equate it to zero.
  • Solve the equation to get the value of x. Let x = a be the critical point of f.
  • Determine the second order derivative of f at x = a.
    • If f’’(a) < 0, x = a is a point of local maxima.
    • If f’’(a) > 0, x = a is a point of local minima.
    • If f’’(a) = 0, then the test fails.

For example, a ball is thrown up in the air. At any instant t, its height is given the equation h = 3 + 14t – 5t2. Let us find the maximum height attained by the ball using the concept of maxima and minima.

The maximum height hmax will be the maximum point of the graph of h = 3 + 14t – 5t2.

Now, h’(t) = 14 – 10t

Equating the above equation to zero, we get

h’(t) = 0

⇒ 14 – 10t = 0

⇒ t = 14/10 = 1.4s

Let us check for maxima and minima at t = 1.4 s.

h’’(t) = –10

∴ h’’(1.4) = –10 < 0.

Thus, t = 1.4 s is a point of maxima and the maximum height attained is hmax = 3 + 14 × 1.4 – 5 (1.4)2 = 12.8 m.

Indefinite Integration

Integration is the reverse process of differentiation. If F is the derivative of f with respect to x, then

∫ F(x) dx = f(x) + C where f’(x) = F(x) and C is the constant of integration.

That is, f(x) is the integral of F. Hence, integration is often referred to as anti-derivatives or primitives.

Formulae for Integration

Following are the formulae we often use while finding the integral of a function. We can compare the integration and differentiation of functions side-by-side for a better understanding.

Derivative

Integral

\(\begin{array}{l}\frac{d}{dx}(c)=0\:where\:c\:is\:a\:constant\end{array} \)

∫ 0 dx = c

\(\begin{array}{l}\frac{d}{dx}(kx)=k\end{array} \)

∫ k dx = kx + c

\(\begin{array}{l}\frac{d}{dx}(sin\;x)=cos\;x\end{array} \)

∫ cos x dx = sin x + c

\(\begin{array}{l}\frac{d}{dx}(cos\;x)=-sin\;x\end{array} \)

∫ sin x dx = –cos x + c

\(\begin{array}{l}\frac{d}{dx}(tan\;x)=sec^{2}\;x\end{array} \)

∫ sec2 x dx = tan x + c

\(\begin{array}{l}\frac{d}{dx}(cot\;x)=-cosec^{2}\;x\end{array} \)

∫ cosec2 x dx = –cot x + c

\(\begin{array}{l}\frac{d}{dx}(sec\;x)=sec\;x\;tan\;x\end{array} \)

∫ sec x tan x dx = sec x + c

\(\begin{array}{l}\frac{d}{dx}(cosec\;x)=-cosec\;x\;cot\;x\end{array} \)

∫ cosec x cot x dx = –cosec x + c

\(\begin{array}{l}\frac{d}{dx}(ln\;x)=\frac{1}{x}\end{array} \)

∫ (1/x) dx = ln x + c

\(\begin{array}{l}\frac{d}{dx}(e^{x})=e^{x}\end{array} \)

∫ ex dx = ex + c

Some rules regarding integration of polynomial functions:

\(\begin{array}{l}1.\:\:\int x^{n}\;dx=\frac{x^{n+1}}{n+1}+C;\;\;n \neq -1\end{array} \)
\(\begin{array}{l}2.\:\:\int \frac{1}{x}\;dx=ln\;x + C;\;\;n = -1\end{array} \)
\(\begin{array}{l}3.\:\:\int kx^{n}\;dx=\frac{kx^{n+1}}{n+1}+ C;\;\;n \neq -1\end{array} \)

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