DifferentiationDifferentiation is the chief chapter in Differential Calculus. It forms the basis of the entire calculus. It also lays the foundation for the subsequent topics of Tangents and Normal and Maxima Minima.

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.

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:

(1) If y = x

^{2}, dy/dx = 2x(2) d (x

^{2}) = 2x

dx

This says that the derivative of x^{2}with respect to x is 2x.(3) If f(x) = x

^{2}, f'(x) = 2x

This says that is f(x) = x^{2}, the derivative of f(x) is 2x.

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. So

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

= {f(a+h) – f(a)} / h

= {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 = dy/ dx, and

= f’ (a)

Therefore dy/dx = f’ (a) = m

m as we’ve seen 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 rules are listed below:

(1)Constant rule:If the function f is a constant, then its derivative is zero.

{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 andβ.

(3)Product Rule:(fg)' = f'g+g'f , for all functions f and g.

(4)Quotient Rule:(f/g)'= (f'g-fg') / g

^{2}, 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 when using the chain rule remember that:

(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 question (i.e. in x's rather than u's).

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

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

(sin x)' = cos x

(cos x)' = -sin x

(tan x)' = sec

^{2}x(Sec x)' = sec x tan x

(cosec x)' = -cosec x cot x

(cot x)' = -cosec

^{2}xThe following video will provide you more clarity

Illustration:What is the gradient of the curve y = 2x

^{3}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 = 6x

^{2}

When x = 3, dy/dx = 6× 9 = 54.

Illustration:

Find the derivative ofWe 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:Find the derivative of f(x) = x

^{4}+ sin (x^{2}) – ln(x)e^{x}+ 7.

Solution:Using the formulae, we get the derivative as

You can also look into the Past Papers to get an idea about the type of questions asked.

Illustration:Find the derivative of:

(x

^{2}+ 5x – 3) / 3 x^{1/2}

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 3x

^{½}.Doing this we obtain

(1/3)x^{3/2}+ (5/3)x^{½}- x^{-½}(using the laws of indices).So differentiating term by term: ½ x

^{½}+ (5/6)x^{-½}+ ½x^{-3/2}.

"Differentiation"is one of the easiest and important chapters of Calculus in the Mathematics syllabus of IIT JEE. The chapter is important not only because it fetches 3-4 questions in most of the engineering examination but also because it is prerequisite to the subsequent chapters of Differential Calculus.

Differentiation is important from the perspective of scoring high in IIT JEE as there are few fixed pattern on which a number Multiple Choice Questions are framed on this topic.