Question

Find the derivative of the following functions from first principle. (i) x^3 – 27 (ii) (x – 1) (x – 2) (iii) 1/x^2 (iv) (x+1) / (x-1)

Answer 1

Given : (i) x^3 – 27 (ii) (x – 1) (x – 2) (iii) 1/x^2 (iv) (x+1) / (x-1)

To find : derivative of the   functions from first principle

Solution:

f'(x) = Lim h→0  (f(x + h) - f(x)) / h

f(x) = x³  - 27

f'(x)  = Lim h→0   ((x + h)³ - x³)/h

=> f'(x)  = Lim h→0    (x³ + h³ + 3x²h + 3xh²  - x³)/h

=> f'(x)  = Lim h→0    ( h³ + 3x²h + 3xh² )/h

=> f'(x)  = Lim h→0    h²+ 3x²  + 3xh

=> f'(x)  =  0 +   3x²  + 0

=> f'(x)  =  3x²

(ii) f(x)  =  (x – 1) (x – 2)

f'(x)  = Lim h→0  ( (x + h - 1)(x + h - 2)  - (x - 1)(x - 2) ) /h

=> f'(x)  = Lim h→0  (  -3x - 3h + 2 + x² + h² + 2xh  - x² +3x - 2) /h

=> f'(x)  = Lim h→0  (   - 3h  + h² + 2xh  ) /h

=> f'(x)  = Lim h→0  (   - 3   + h  + 2x   )  

=> f'(x)  =  -3 + 0 + 2x

=> f'(x)  = 2x - 3

f(x)  =  1/x^2

f'(x)  = Lim h→0     (1/(x + h)²  - 1/x²)/h

=> f'(x)  = Lim h→0   ( x²  - (x + h)²)/ (x + h)²x²h

=>  f'(x)  = Lim h→0   ( -h² - 2xh )/ (x + h)²x²h

=>  f'(x)  = Lim h→0   ( -h  - 2x  )/ (x + h)²x²

=>  f'(x)  =  ( 0  - 2x  )/ (x + 0)²x²

=>  f'(x)  =   - 2/x³

f(x)  =  (x+1) / (x-1)

f'(x)  = Lim h→0      ((x + h + 1)/(x + h - 1)  - (x + 1)/(x - 1) )/h

=> f'(x)  = Lim h→0   ( x² + xh + x - x - h - 1 - (x² + xh - x + x + h - 1) ) /((x + h - 1) (x - 1) )h

=> f'(x)  = Lim h→0   (   -2h ) /((x + h - 1) (x - 1) )h

=> f'(x)  = Lim h→0   (   -2  ) /((x + h - 1) (x - 1) )

=> f'(x)  =    (   -2  ) /((x   - 1) (x - 1) )

=>  f'(x)  =    -2/(x - 1)²

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