f(x)=∫0x dt1+t3 and g(x) be the inverse of f(x). Then the value of 4g′′(x)(g(x))2 is __.

Solution:

$y = f (x) \Rightarrow x = f^{- 1} (y) \Rightarrow x = g (y)$

Given $y = f (x) = \int_{0}^{x} \frac{d t}{\sqrt{1 + t^{3}}}$

$\frac{d y}{d x} = \frac{1}{\sqrt{1 + x^{3}}}$ or $\frac{d x}{d y} = \sqrt{1 + x^{3}}$

$\begin{array}{l} g^{'} (y) = \sqrt{1 + g^{3} (y)} \\ g^{''} (y) = \frac{3 g^{2} (y) g^{'} (y)}{2 \sqrt{1 + g^{3} (y)}} \\ ∴ 2 g^{''} (y) = 3 g^{2} (y) \frac{g^{'} (y)}{\sqrt{1 + g^{3} (y)}} = 3 g^{2} (y) \frac{\sqrt{1 + g^{3} (y)}}{\sqrt{1 + g^{3} (y)}} = 3 g^{2} (y) \end{array}$

or $2 g^{''} (y) = 3 g^{2} (y)$

$f (x) = \int_{0}^{x} \frac{d t}{\sqrt{1 + t^{3}}}$ and g(x) be the inverse of $f (x)$ . Then the value of $4 \frac{g^{''} (x)}{(g (x))^{2}}$ is __.

Solution:

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