2 - Power Rule for Differentiation (1)

Let y=f(x)= x n  , where n  is a positive integer. Then

dy dx = lim h0 f(x+h)f(x) h

= lim h0 (x+h) n x n h

= lim h0 [(x+h)x] [ (x+h) n1 + (x+h) n2 x+ (x+h) n3 x 2 ++(x+h) x n2 + x n1 ] h

= lim h0 h[ (x+h ) n1 + (x+h ) n2 x+ (x+h ) n3 x 2 ++ (x+h ) x n2 + x n1 ] h

= lim h0 [ (x+h ) n1 + (x+h ) n2 x+ (x+h) n3 x 2 ++ (x+h ) x n2 + x n1 ]

=n x n1

If y=f(x)=k  , where k  is a constant, then

dy dx = lim h0 f(x+h)f(x) h

= lim h0 kk h

= lim h0 0 h

= lim h0 0

=0

In particular, if y=f(x)=1= x 0  , then

dy dx =0=0 x 01

Thus we see that if y=f(x)= x n  , where n  is a nonnegative integer, then

dy dx =n x n1