Let g is the inverse function of f and f-x-x101-x2- If g2-a then g-2 is equal to

The correct option is B 1+a^2a^10 g is inverse of function f f'(x) = x^10/(1+x^2) g(2) = a g'(2) = (1) → g is inverse of function f, th ...

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Differentiation of Inverse Trigonometric Functions

Let g is th...

Question

Let $g$ is the inverse function of $f$ and $f^{'} (x) = \frac{x^{10}}{(1 + x^{2})}$ . If $g (2) = a$ then $g^{'} (2)$ is equal to

\frac{5}{2^{10}}

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\frac{1 + a^{2}}{a^{10}}

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\frac{a^{10}}{1 + a^{2}}

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\frac{1 + a^{10}}{a^{2}}

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Solution

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Similar questions

g

f

f^{'} (x) = \frac{x^{10}}{1 + x^{2}} .

g (2) = a,

g^{'} (2)

{(1 + x)}^{10} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . a_{10} x^{10}

{(a_{0} - a_{2} + a_{4} - a_{6} + a_{8} - a_{10})}^{2} + {(a_{1} - a_{3} + a_{5} - a_{7} + a_{9})}^{2}

a_{1}, a_{2}, a_{3}, \dots

\frac{a_{1} + a_{2} + \dots + a_{10}}{a_{1} + a_{2} + \dots + a_{p}} = \frac{100}{p^{2}},

p \neq 10,

\frac{a_{11}}{a_{10}}

{(1 + x + 2 x^{2})}^{20} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . + a_{10} x^{40}

a_{1} + a_{3} + a_{5} + . . . . . . + a_{39}

a, a_{1}, a_{2} . . . . . . . . . . . . . a_{10}, b

a, g_{1}, g_{2} . . . . . . g_{10}, b

h

a

b

\frac{a_{1} + a_{2} + . . . . . . . . . + a_{10}}{g_{1} g_{10}} + \frac{a_{2} + a_{3} + . . . . . . . . . + a_{9}}{g_{2} g_{9}} + . . . . . . . . . . . + \frac{a_{5} + a_{6}}{g_{5} g_{6}}

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