Let f:(-1,1)R be a differentiable function with f(0)=-1 and f'(0)=1.

Question

Let $f:(-1,1)	o\mathbb{R}$ be a differentiable function with $f(0)=-1$ and $f'(0)=1$. Let $g(x)=[f(2f(x)+2)]^2$. Then $g'(0)=$

Accepted Answer

$$g'\left( x \right) = 2\left( {f\left( {2f\left( x \right) + 2} \right)} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{d \over {dx}}\left( {f\left( {2f\left( x \right) + 2} \right)} \right)} \right)$$

$$ = 2f\left( {2f\left( x \right) + 2} \right)f'\left( {2f\left( x \right)} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left. 2 \right).\left( {2f'\left( x \right)} \right)$$

$$ \Rightarrow g'\left( 0 \right) = 2f\left( {2f\left( 0 \right) + 2} \right).$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( {2f\left( 0 \right) + 2} \right).2f'\left( 0 \right)$$

$$ = 4f\left( 0 \right){\left( {f'\left( 0 \right)} \right)^2}$$

$$ = 4\left( { - 1} \right){\left( 1 \right)^2} =  - 4$$

Solutions · 1