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