DIFFERENTIATION FROM FIRST PRINCIPLES

Given

y  =  f(x)

its derivative, or rate of change of y with respect to x is defined as

Example 1 :

Differentiate xfrom first principles.

Solution :

f'(x)  =  lim h-> 0   [f(x+h) - f(x)]/h  ---(1)

f(x)  =  x2

f(x+h)  =  (x+h)2

f(x+h)  =  x2+2hx+h2 

f(x+h)-f(x)  =  x2+2hx+h2 - x2

f(x+h)-f(x)  =  2hx+h2

By applying the above value in the formula, we get

f'(x2)  =  lim h-> 0   [2hx+h2]/h

f'(x2)  =  lim h-> 0   [2x+h]

By applying the limit, we get

f'(x2)  =  2x

Example 2 :

Differentiate 2x2-x from first principles when x  =  3. 

Solution :

f'(x)  =  lim h-> 0   [f(x+h) - f(x)]/h  ---(1)

f(x)  =  2x2-x

f(x+h)  =  2(x+h)2-(x+h)

f(x+h)  =  2(x2+2hx+h2)-(x+h)

f(x+h)  =  2x2+4hx+2h2-x-h

f(x+h)-f(x)  =  2x2+4hx+2h2-x-h-(2x2-x)

=  2x2+4hx+2h2-x-2x2+x-h

f(x+h)-f(x)  =  4hx+2h2-h

By applying the above value in the formula, we get

f'(2x2-x)  =  lim h-> 0   [4hx+2h2-h]/h

f'(2x2-x)  =  lim h-> 0   (4x+2h-1)

By applying the limit, we get

f'(2x2-x)  =  4x-1

When x  =  3,

=  4(3)-1

=  11

=  12

So, differentiation of 2x2-x, when x  =  3 is 12.

Example 3 :

Differentiate 1/x from first principles.

Solution :

f'(x)  =  lim h-> 0   [f(x+h) - f(x)]/h  ---(1)

f(x)  =  1/x

f(x+h)  =  1/(x+h)

f(x+h) - f(x)  =  1/(x+h) - (1/x)

=  x-(x+h)/x(x+h)

f(x+h) - f(x)  =  -h/x(x+h)

By applying the above value in the formula, we get

f'(1/x)  =  lim h-> 0   [-h/x(x+h)]/h

=  lim h-> 0   [-1/x(x+h)]

By applying the limit, we get

=  [-1/x(x)]

=  -1/x2

Example 4 :

Differentiate log x from first principles.

Solution :

f'(x)  =  lim h-> 0   [f(x+h) - f(x)]/h  ---(1)

f(x)  =  logx

f(x+h)  =  log(x+h)

f(x+h) - f(x)  =  log(x+h) - logx

=  log[(x+h)/x]

=  log(1+(h/x))

By applying the above value in the formula, we get

f'(log x)  =  lim h-> 0   [log(1+(h/x))]/h

By multiplying the denominator by x/x.

f'(log x)  =  lim h-> 0   [log(1+(h/x))]/x(h/x)

=  (1/x) lim h-> 0   [log(1+(h/x))]/(h/x)

=  1/x

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