Denition approach of learning new topics

21
dy
dx = lim f(x + h) − f(x)
h→0 h
=
(x + h) 2 − x 2
lim
h→0 h
=
2xh + h 2
lim
h→0 h
=
2x + h
lim
h→0 1
(since h → 0, hence h ≠ 0)
= 2x
Problem. Find the derivative of f(x) = x n where n ∈ N
Again using the denition of derivative
dy
dx = lim f(x + h) − f(x)
h→0 h
(x + h) n − x n
= lim
h→0 h
= lim
h→0
h(nx n−1 + c 1 x n−2 h + c 2 x n−3 h 2 + . . . + h n−1 )
h
= nx n−1
Hence we can write
d
dx (xn ) = nx n−1 . Now we can nd any of these
d
dx (x2 ) = 2x
d
dx (x3 ) = 3x 2
d
dx (x4 ) = 4x 3 . So we have ended up nding general formula for derivative of
functions of the form variable constant
Problem. Find the derivative of f(x) = 2
This is a constant function.
dy
dx = lim f(x + h) − f(x)
h→0 h
2 − 2
= lim
h→0 h
= 0
Can you see that derivative of a constant function is coming out to be zero.
That means derivative of any function f(x) = c is Zero.
Problem. Prove the relation d
d
(k · f(x)) = k · (f(x)) where k is a constant.
dx dx
What we mean here is a constant can come out of the derivative operator.
By denition,