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If f : R → R is a function defined as f x 3 = x 5 ,   ∀ x ∈ R - 0 and f x is differentiable ∀ x ∈ R , then the value of 1 4 f ' 27 is equal to (here f ' represents the derivative of f )

Correct answer

3.75

Step-by-step solution

f x 3 = x 5 On differentiating with respect to x f ' x 3 ⋅ 3 x 2 = 5 ⋅ x 4 f ′ x 3 = 5 3 x 2 Putting x = 3 , we get, f ′ 27 = 5 3 9 = 15

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