if-f-o-g-x-x-and-f-x-1-f-x-2-then-g-2-

Question Number 106039 by 175mohamed last updated on 02/Aug/20

i f (f o g) (x) = x a n d f^{'} (x) = 1 + {(f (x))}^{2}

t h e n g^{'} (2) = \dots .

Answered by bemath last updated on 02/Aug/20

\frac{d (f \circ g) (x)}{dx} = g^{'} (x) . f^{'} (g (x)) = 1

g^{'} (2) . f^{'} (g (2)) = 1

g^{'} (2) = \frac{1}{1 + (f {(g (2))}^{2}}

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

so g (2) = f^{- 1} (2)

∴ g^{'} (2) = \frac{1}{1 + (f {(f^{- 1} (2))}^{2}} = \frac{1}{1 + 4} = \frac{1}{5}

Answered by maouame last updated on 02/Aug/20

i f (f o g) (x) = x a n d f^{'} (x) = 1 + {(f (x))}^{2}

t h e n g^{'} (2) = \dots .

O n a {(f o g)}^{'} (x) = g^{'} (x) \times f^{'} (g (x))

A l o r s g^{'} (x) \times f^{'} (g (x)) = 1

\Rightarrow g^{'} (x) = \frac{1}{f^{'} (g (x))} (c a r f^{'} (x) > 0)

\Rightarrow g^{'} (x) = \frac{1}{1 + {(f (g (x)))}^{2}}

\Rightarrow g^{'} (x) = \frac{1}{1 + x^{2}}

\Rightarrow g^{'} (2) = \frac{1}{5}

Answered by mathmax by abdo last updated on 02/Aug/20

fog (x) = x \Rightarrow f^{^{'}} (g (x)) . g^{^{'}} (x) = 1 but f^{^{'}} (x) = 1 + {(f (x))}^{2} \Rightarrow

f^{^{'}} (g (x)) = 1 + {(fog (x))}^{2} = 1 + x^{2} \Rightarrow f^{^{'}} (g (2)) = 1 + 2^{2} = 5 and

f^{^{'}} (g (2)) . g^{^{'}} (2) = 1 \Rightarrow g^{^{'}} (2) = \frac{1}{5}