Continuity and DifferentiabilityMedium
Question
Let f be a function defined for all x ∈ R. If f is differentiable and f(x3) = x5 for all x ∈ R ( x ≠ 0), then the value of f′(27) is -
Options
A.15
B.45
C.0
D.None
Solution
f(x3) = x5
find f′(27) = ?
f′(x3) 3x2 = 5x4 Put x = 3
f′(27) × 3 × 9 = 5 × 27 × 3
find f′(27) = ?
f′(x3) 3x2 = 5x4 Put x = 3
f′(27) × 3 × 9 = 5 × 27 × 3
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