# HOW TO FIND DERIVATIVES USING FIRST PRINCIPLE

How to Find Derivatives Using First Principle :

Here we are going to see how to find derivatives using first principle

Let f be defined on an open interval I ⊆ R containing the point x0, and suppose that

exists. Then f is said to be differentiable at x0 and the derivative of f at x0, denoted by f'(x0) , is given by

For a function y = f(x) defined in an open interval (a, b) containing the point x0, the left hand and right hand derivatives of f at x = h are respectively denoted by f'(h-) and f'(h+)

f'(h-)  =  lim h-> 0-[f(x + h) - f(x)] / h

f'(h+)  =  lim h-> 0+[f(x + h) - f(x)] / h

provided the limits exist.

Question 1 :

Find the derivatives of the following functions using first principle.

Solution :

(i) f(x)  =  6

f'(x)  =  lim h-> 0[f(x + h) - f(x)] / h

f(x + h)  =  6

f'(x)  =  lim h-> 0((6) - 6)/h

=  lim h-> 0  (0/h)

=  0

(ii)  f(x)  =  -4x + 7

f'(x)  =  lim h-> 0[f(x + h) - f(x)] / h

f(x + h)  =  -4(x + h) + 7

f'(x)  =  lim h-> 0(-4(x + h) + 7 - (-4x + 7))/h

=  lim h-> 0(-4x - 4h + 7 + 4x - 7)/h

=  lim h-> 0  (-4h/h)

=  -4

(ii)  f(x)  =  -x2 + 2

f'(x)  =  lim h-> 0[f(x + h) - f(x)] / h

f(x + h)  =  -(x + h)2 + 2

=  - (x2 + h2 + 2xh) + 2

=  - x2 - h2 - 2xh + 2

f'(x)  =  lim h-> 0((- x2 - h2 - 2xh + 2) - (-x2 + 2))/h

=  lim h-> 0(- x2 - h2 - 2xh + 2 + x2 - 2)/h

=  lim h-> 0(- h2 - 2xh)/h

=  lim h-> 0(- h - 2x)

=  -2x

Question 2 :

Find the derivatives from the left and from the right at x = 1 (if they exist) of the following functions. Are the functions differentiable at x = 1?

(i)  f(x)  =  |x - 1|

Solution :

If the function is differentiable, then

f'(1-)  =  f'(1+)

f'(1-)  =  [f(x) - f(1)] / (x - 1)

=  [-(x - 1) - 0]/(x - 1)

=  -1

f'(1+)  =  [f(x) - f(1)] / (x - 1)

=  [(x - 1) - 0]/(x - 1)

=  1

Hence the given function is not differentiable at x = 1.

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