Tính đạo hàm của g(x)=f^2(x^2+2x+m)

Đáp án: $g'\left( x \right) = \left( {4x + 4} \right).f'\left( {{x^2} + 2x + m} \right).f\left( {{x^2} + 2x + m} \right)$

Giải thích các bước giải:

$\begin{array}{l}
g\left( x \right) = {f^2}\left( {{x^2} + 2x + m} \right)\\
 \Leftrightarrow g'\left( x \right) = 2.\left( {{x^2} + 2x + m} \right)'\\
.f'\left( {{x^2} + 2x + m} \right).f\left( {{x^2} + 2x + m} \right)\\
 \Leftrightarrow g'\left( x \right) = 2.\left( {2x + 2} \right).f'\left( {{x^2} + 2x + m} \right).f\left( {{x^2} + 2x + m} \right)\\
 \Leftrightarrow g'\left( x \right) = \left( {4x + 4} \right).f'\left( {{x^2} + 2x + m} \right).f\left( {{x^2} + 2x + m} \right)
\end{array}$

$g(x)=f^2(x^2+2x+m)$

$g'(x)=2f(x^2+2x+m).[f(x^2+2x+m)]'$

$=2f(x^2+2x+m).(x^2+2x+m)'.f'(x^2+2x+m)$

$=2(2x+2)f'(x^2+2x+m).f(x^2+2x+m)$