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unit 1 differential calculus - IGNOU

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1.6.1 Derivative of a Function<br />

We now define the derivative of a given function by using the notion of limit.<br />

Definition 10<br />

The function y = f (x) is said to have a derivative (or be differentiable) at a<br />

point x if, and only if, the limit<br />

lim<br />

h →0<br />

f ( x + h)<br />

− f ( x)<br />

h<br />

dy<br />

exists and is finite. This limit, written as or f ′ (x), is called the<br />

dx<br />

derivative of f (x) with respect to x.<br />

If the derivative of a function exists at every point of an interval, then we say that<br />

the function is differentiable in that interval. However, while considering the<br />

derivatives at the end points of an interval, we evaluate suitable one-sided<br />

derivatives. For example, if the interval is [a, b], then the derivative at x = a is<br />

calculated as<br />

lim<br />

+<br />

h →0<br />

f ( a + h)<br />

− f ( a)<br />

and at b as<br />

h<br />

lim<br />

−<br />

h →0<br />

f ( b + h)<br />

− f ( b)<br />

h<br />

If f ′ (x) is continuous at a point, then we say that f (x) is continuously<br />

differentiable at that point.<br />

Example 1.21<br />

Solution<br />

Let y = x n (n is a positive integer). Prove that .<br />

1 − dy n<br />

= nx<br />

dx<br />

Then<br />

y = f ( x)<br />

=<br />

n<br />

x<br />

n<br />

f ( x + h)<br />

− f ( x)<br />

( x + h)<br />

− x<br />

=<br />

h<br />

h<br />

( x + h − x)<br />

−1<br />

n −<br />

=<br />

[ ( x + h)<br />

+ ( x + h)<br />

h<br />

n 2<br />

The above step is derived by using the result<br />

We now have<br />

f<br />

a<br />

n<br />

− b<br />

n<br />

= ( a − b)<br />

[ a<br />

x<br />

n<br />

+<br />

. . .<br />

+ ( x + h)<br />

x<br />

b + . . . + ab<br />

n − 2 n −1<br />

+ x<br />

n −1<br />

n − 2<br />

n − 2 n −1<br />

+ a<br />

+ b<br />

+ −<br />

n<br />

= ( x + h)<br />

+ ( x + h)<br />

x + . . . + ( x + h)<br />

x + x<br />

h<br />

( x h)<br />

f ( x)<br />

n −1<br />

n − 2<br />

n − 2 −1<br />

Taking the limit h → 0, we get<br />

dy<br />

dx<br />

=<br />

lim<br />

h→<br />

0<br />

f ( x + h)<br />

− f ( x)<br />

= x<br />

h<br />

Therefore, .<br />

1 − dy n<br />

= nx<br />

dx<br />

n −1<br />

+ x<br />

n − 2<br />

x +<br />

. . .<br />

+ x<br />

n −1<br />

(n number of terms)<br />

]<br />

]<br />

Differential Calculus<br />

37

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