Question

Find the derivative of the function given by f(x) = (1 + x) (1 + x²) (1 + x⁴) (1 + x⁸) and hence find f′(1).

Answer 1

Given, f(x) = (1 + x) (1 + x²) (1 + x⁴) (1 + x⁸)

Taking logarithm of both sides,

log f(x) = log [(1 + x) (1 + x²) (1 + x⁴) (1 + x⁸)]

log f(x) = log (1 + x) + log (1 + x²) + log (1 + x⁴) + log (1 + x⁸)   ...[∵ log mn = log m + log n]

Differentiating both sides with respect to x,

`1/(f (x)) d/dx f(x) = 1/(1 + x) d/dx (1 + x) + 1/(1 + x^2) d/dx (1 + x^2) + 1/(1 + x^4) d/dx (1 + x^4) + 1/(1 + x^8) d/dx (1 + x^8)`

or `f'(x) = 1/(1 + x) + (2x)/(1 + x^2) + (4x)/(1 + x^4) + (8x)/(1 + x^8)`

or `f'(x) =  f (x) [1/(1 + x) + (2x)/(1 + x^2) + (4x^3)/(1 + x^4) + (8x^7)/(1 + x^8)]`

= `(1 + x) (1 + x^2) (1 + x^4)(1 + x^8) [1/(1 + x) + (2x)/(1 + x^2) + (4x^3)/(1 + x^4) + (8x^7)/(1 + x^8)]`

Putting x = 1,

f'(1) = (1 + 1) (1 + 1) (1 + 1) (1 + 1) `xx [1/(1 + 1) + 2/(1 + 1) + 3/(1 + 1) + 4/(1 + 1)]`

= `2 xx 2 xx 2xx 2 xx [1/2 + 2/2 + 4/2 + 8/2]`

= `(2 xx 2 xx 2xx 2)/2 [1 + 2 + 4 + 8]`

= 8 × 15

= 120