Let f : R → R be a function such that f '(x) = 5x4 + 3x2 + 5 and f (1) + f (–1) = 0. If g(f(x)) = x and h(x) = g(x2 + 4x

Engineering

Mathematics

Derivative of Inverse Function

Question

Let f : R → R be a function such that f '(x) = 5x⁴ + 3x² + 5 and f (1) + f (–1) = 0. If g(f(x)) = x and h(x) = g(x² + 4x – 5), then

h'(2) = $\frac{8}{13}$

h'(– 6) = $\frac{- 8}{13}$

h'(– 5) = $\frac{6}{5}$

h'(1) = $\frac{- 6}{5}$

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f '(x) = 5x⁴ + 3x² + 5
f (x) = x⁵ + x³ + 5x + c
f (1) + f (–1) = 0 ⇒ c = 0
$∴$ f (x) = x⁵ + x³ + 5x
g(f(x)) = x ⇒ g is the inverse function of f
Now,   h(x) = g(x² + 4x – 5)
h'(x) = g'(x² + 4x – 5) (2x + 4)
(A)   h'(– 6) = g'(7) (–8) = $\frac{- 8}{f' (1)} = \frac{- 8}{13}$        { $∵$ f(1) = 7}
(B)   h'(2) = g'(7) (8) = $\frac{8}{f' (1)} = \frac{8}{13}$
(C)   h'(–5) = g'(0) (– 6) = $\frac{- 6}{f' (0)} = \frac{- 6}{5}$        { $∵$ f(0) = 0}
(D)   h'(1) = g'(0) (6) = $\frac{6}{f' (0)} = \frac{6}{5}$