Differentiation

Differentiation is the chief chapter in Differential Calculus. In fact, it may be termed as the essence of calculus as it forms the basis of the entire calculus. It also lays the foundation for the subsequent topics like Tangents and Normal and Maxima Minima. In this section, 

Normally a dependent variable is expressed in terms of independent variables by means of an equation. Now when we find the differential coefficient of the dependent variable with respect to the independent variable, what we do is we try to find out another equation by which the change in the dependent variable (for any infinitesimal change in independent variable) is relatable to the independent variable, whatever be the value of the independent variable.

Hence, derivative is a measure of how a function changes as its input changes. Informally, the derivative is the ratio of the infinitesimal change of the output over the infinitesimal change of the input producing that change of output. Geometrically, for a real valued function of a single real variable, the derivative at a point equals the slope of the tangent line to the graph of the function at that point. The process of finding a derivative is called differentiation.

In fact, differentiation and integration are the two fundamental operations in single-variable calculus. The figure given below illustrates the exact difference between integration and differentiation:

Notation:

There are a number of ways of writing the derivative of a function. Though all these are same but it is essential to know them so as to avoid any kind of confusion:

Suppose we are finding the derivative of x2, then its derivative may be written as:

  • If y = x2, dy/dx = 2x

  • d/dx (x2) = 2x

This says that the derivative of x2 with respect to x is 2x.

  • If f(x) = x2, f'(x) = 2x

This says that is f(x) = x2, the derivative of f(x) is 2x.

Geometrical Interpretation of Differentiation

Let y = f(x) be a function, and let P(a, f(a)) and Q(a+h, f(a+h)) be two points on the graph of the function that are close to each other. This graph is shown below:

Geometrical Interpretation of Differentiation
Joining the points P and Q with a straight line gives us the secant on the graph of the function. And in the ∆PQR, the gradient of the line is given by

m = {f(a+h) – f(a)}/(a+h-a) = {f(a+h) – f(a)} / h

In limiting process, i.e. as Q approaches P, h becomes really small, almost close to zero.

Hence, m = (change in y)/(change in x) 

                = {f(a+h) – f(a)} / h

 lim h→0 (∆y/∆x) = {f(a+h) – f(a)} / h

As Q → P, the chord/secant PQ tends to be a tangent at P for the curve y = f(x). Thus the limiting becomes the slope of the tangent at P for y = f(x).

We denote lim h→0 (∆y/∆x) = dy/dx, and

lim h→0 [f(a+h) – f(a)]/ h = f’(a) 

Therefore dy/dx = f’ (a) = m

Now, we have already seen that ‘m’ is the slope or gradient of the tangent at P for f(x).

f’(a)  =  tan θ  = slope of f(x) at x = a.

Rules of differentiation:

The differentiation of functions is carried out in accordance with some rules. These differentiation rules have been listed with the help of the following chart:

Rules of Differentiation

All these rules will be discussed in detail in the coming sections. Here, we shall give a brief outline of these rules:

(1) Constant rule:

If the function f is a constant, then its derivative is zero. Moreover if a function is multiplied by a constant then its derivative is given by 

{cf(x)}' = cf'(X)

(2)  Sum Rule:

{f(x) ± g(x)}' = f'(x) ± g'(x)

(αf + βg)' = αf' + βg' , for all functions f and g and real numbers α and β.

(3) Product Rule:

(fg)' = f'g + g'f, for all functions f and g.

(4) Quotient Rule:

(f/g)' = (f'g – fg')/g2, for all functions f and g such that g ≠ 0.

(5) Chain Rule:

If f(x) = h(g(x)), then f'(x) = h'(g(x)). g'(x)

So while using the chain rule remember the following points:

(i) Express the original function as a simpler function of u, where u is a function of x.

(ii) Differentiate the two functions you now have. Multiply the derivatives together, leaving your answer in terms of the original function (i.e. in x's rather than u's).

Logarithmic Differentiation

In order to compute the derivative of:

  • A function which is the product or quotient of a number of functions

y = g1(x)g2(x)g3(x) ….. or [(g1(x) g2(x) g3(x) ….) / (h1(x) h2(x) h3(x) ….)

  • A function of the form [f(x)]g(x) where f and g are both derivable, it is better to take the logarithm of the function first and then differentiate.

  • Or, we can write it as y = (f(x))g(x) = e g(x).ln f(x)

​We list below certain differentiation formulae for finding the derivative of functions:

Differentiation Formulae

Given below are some more formulae of some logarithmic and trigonometric functions:

d/dx (xx) = xx (1 + ln x)

d/dx (logax) = 1/ x ln a

d/dx (logf(x)) = 1/ f(x) ln a . d/dx f(x)

d/dx (sin x) = (sin x)' = cos x

(cos x)' = -sin x

(tan x)' = sec2 x

(sec x)' = sec x tan x

(cosec x)' = -cosec x cot x

(cot x)' = -cosec2x

We now discuss some of the differentiation examples followed by some differentiation questions:

Illustration 1:

What is the gradient of the curve y = 2x3 at the point (3, 54)?

Solution:

The gradient of the curve is given by its derivative so the question actually requires you to compute the derivative.

The derivative is dy/dx = 6x2

When x = 3, dy/dx = 6 × 9 = 54.

Hence, the required gradient is 54.

________________________________________________________________________________

Illustration 2:

Find the derivative of  

Solution:

We will use the quotient rule to find the derivative.

Now we know that

 ……….(1)

  …….(2)

Hence, combining (1) and (2) we get the required answer.

_______________________________________________________________________________

Illustration 3:

Find the derivative of f(x) = x4 + sin (x2) – ln (x)ex + 7.

Solution: 

Using the formulae, we get the derivative as

___________________________________________________________________________-

Illustration 4: 

Find the derivative of (x2 + 5x – 3)/ 3 x½.

Solution:

This looks hard, but it isn't. The trick is to simplify the expression first: do the division (divide each term on the numerator by 3x½.

Doing this we obtain

(1/3)x3/2 + (5/3)x½ - x (using the laws of indices).

So differentiating term by term: ½ x½ + (5/6)x + ½x-3/2.


Q1. Both the derivative and the integral of this function are same

(a) the function is ex.

(b) the function is e2x.

(c) the function is 2ex.

(d) the function is ln x.

Q2. If f(x) = h(g(x)), then f'(x) =

(a) h'(g’(x)). g(x)

(b) h(g’(x)). g'(x)

(c) h'(g(x)). g(x)

(d) h'(g(x)). g'(x)

Q3. The derivative of logax is 

(a) 1/x ln ax

(b) 1/x ln ax

(c) 1/x ln ax

(d) 1/x ln ax

Q4. The derivative of a constant function is 

(a) zero

(b) the function itself.

(c) constant

(d) none of these

Q5. the derivative of the function [f(x)]n is

(a) n f(x)n (f’(x))

(b) n f’(x)n-1 (f(x))

(c) n f(x)n-1 (f’(x))

(d) f(x)n-1 (f’(x))

Q1.

Q2.

Q3.

Q4.

Q5.

(a)

(d)

(b)

(a)

(c)

 

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