Let f be a real-valued function defined on the interval (–1, 1) such that , for all x ∈ (-1,1), and let f–1 be the inverse function of f.Then (f

Engineering

Mathematics

Leibnitz Rule Derivative of Anti Derivatives

Question

Let f be a real-valued function defined on the interval (–1, 1) such that $e^{- x} f (x) = 2 + \int_{0}^{x} \sqrt{t^{4} + 1} dt$ , for all x ∈ (-1,1), and let f^–1 be the inverse function of f.

Then (f^–1)' (2) is equal to

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{e}$

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Let f^–1(x) = g (x) then g[f (x)] = x $\Rightarrow$ g ' [f (x)] f ' (x) = 1

at x = 0 f (x) = 2 so put x = 0 to get

g ' [f (0)] f ' (0) = 1

also $f' (x) = e^{x} [2 + \int_{0}^{x} \sqrt{1 + t^{4}} dt] + e^{x} (1 + x^{4}) \Rightarrow$ f ' (0) = 3

Hence g ' (2) = $\frac{1}{3}$

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