Given-that-f-and-g-are-differentiable-functions-such-that-f-x-1-1-f-x-2-and-g-f-1-find-g-x-

Question Number 153963 by ZiYangLee last updated on 12/Sep/21

Given that f and g are differentiable

functions such that f^{'} (x) = \frac{1}{\sqrt{1 + {(f (x))}^{2}}},

and g = f^{- 1}, find g^{'} (x) .

Answered by mr W last updated on 12/Sep/21

y = f (x)

y^{'} = \frac{1}{\sqrt{1 + y^{2}}}

\sqrt{1 + y^{2}} d y = d x

\frac{1}{2} [\ln (y + \sqrt{1 + y^{2}}) + y \sqrt{1 + y^{2}}] + C = x

\Rightarrow g (x) = f^{- 1} (x) = \frac{1}{2} [\ln (x + \sqrt{1 + x^{2}}) + x \sqrt{1 + x^{2}}] + C

g^{'} (x) = \frac{1}{2} [\frac{1 + \frac{x}{\sqrt{1 + x^{2}}}}{x + \sqrt{1 + x^{2}}} + \sqrt{1 + x^{2}} + \frac{x^{2}}{\sqrt{1 + x^{2}}}]

= \frac{1}{2} [\frac{x + \sqrt{1 + x^{2}}}{x + \sqrt{1 + x^{2}}} + 1 + 2 x^{2}] \frac{1}{\sqrt{1 + x^{2}}}

= \sqrt{1 + x^{2}}

Commented by ZiYangLee last updated on 12/Sep/21

i found a better way . .

Commented by mr W last updated on 12/Sep/21

y e s, y o u r w a y i s b e s t!

Answered by ZiYangLee last updated on 12/Sep/21

l e t g (x) = f^{- 1} (x) = y

f (y) = x

f^{'} (x) \frac{d y}{d x} = 1

\frac{1}{\sqrt{1 + {[f (y)]}^{2}}} \frac{d y}{d x} = 1

\frac{d y}{d x} = \sqrt{1 + {[f (y)]}^{2}}

g^{'} (x) = \sqrt{1 + {[f (y)]}^{2}}

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