Given g(x) | = | [f(2 f(x) + 2)]2 |
⇒ g'(x) | = | 2 f(2 f(x) + 2)).f '(2 f(x) + 2).2f '(x) |
Substituting x | = | 0, |
⇒ g'(0) | = | 2f(2 f(0) + 2)).f '(2f(0) + 2).2f '(0) |
= | 2f(0).f '(0).2f '(0) (∵ f(0) = –1 and f '(0) = 1) | |
= | 2(–1)(1).2(1) = – 4 |
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