Continuity and DifferentiabilityHard
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
Create a free account to view solution
View Solution FreeMore Continuity and Differentiability Questions
If g(x) = xtan-1x, then the value of g′(1) equals -...If f(x) = , then correct statement is-...The derivative of f(tanx) w.r.t. g(secx) at x = π/4 , where f′(1) = 2 and g′ (√2 ) = 4, is -...If f (x) = , then f′(1) equals -...The number of points where f(x) = [sin x + cos x] (where [ ] denotes the greatest integer function), x ∈ (0, 2`...