If $f:( - 1,1) \rightarrow R$ be a differentiable function with $f(0) = - 1$ and $f^{'}(0) = 1.$ Let g($x) = \left\lbrack f\left( 2f(x) + 2 \right) \right\rbrack^{2}$. Then, $g^{'}(0)$ is equal to

Question

Dr. Abubacker · Accepted Answer

(b)
G$iven\ ,g(x) = \lbrack f{\left( 2f(x) + 2 ight)brack}^{2}$
$	herefore g^{'}(x) = 2.f\left( 2f(x) + 2 ight).f^{'}\left( 2f(x) + 2 ight).2f^{'}(x)$
$= 4\ f\left( 2f(x) + 2 ight)f^{'}\left( 2f(x) + 2 ight)f^{'}(x)$
$	herefore g^{'}(0) = 4f(0)f^{'}(0)f^{'}(0) = - 4$

Q. If $f:( - 1,1) \rightarrow R$ be a differentiable function with $f(0) = - 1$ and $f^{'}(0) = 1.$ Let g($x) = \left\lbrack f\left( 2f(x) + 2 \right) \right\rbrack^{2}$. Then, $g^{'}(0)$ is equal to

Explanation

Reviewed by Dr. Abubacker

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