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Question

Find the derivative of the following functions from first principle. 

(i) x3 - 27
(ii) (x-1) (x-2)

(iii) 1x2 
(iv) x+1x1


Solution

(i) Here f(x) = x3 - 27 

Then f(x+h) = (x+h)3 - 27 

We know that f' = limh0f(x+h)f(x)h

f' (x) = limh0(x+h)327x3+27h

= limh0(x+h)327x3+27h

= limh0h(h2+3x2+3xh)h=3x2 

(ii) Here f(x) = (x-1) (x-2) = x2 - 3x+2

Then f(x+h) = (x+h-1) (x+h-2)

= x2+h2 + 2xh - 3x - 3h + 2

We know that f'(x) = limh0f(x+h)f(x)h

limh0h(h+2x3)h = 2x - 3

(iii) Here f(x) = 1x2

Then f(x+h) = 1(x+h)2

We know that f'(x) = limx0f(x+h)f(x)h

f'(x) =limx01(x+h)21x2h

=limh0x2(x+h)2hx2(x+h)2

=limh0x2x2h22xhhx2(x+h)2

=limh0h(h2x)hx2(x+h)2

= 2xx2×x2=2x3.

(iv) Here f(x) = x+1x1

Then f(x+h) = x+h+1x+h1

We know that f'(x) = limh0f(x+h)f(x)h

f'(x) = limh0x+h+1xh1x+1x1h

= limh0(x+h+1)(x1)(x+1)(x+h1)h(x+h1)(x1)

= limh0x2x+xhh+x1x2xh+xxh+1h(x+h1)(x1)

= limh02hh(x+h1)(x1)

= 2(x1)2.
  

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