1,cho hs f(x) = log2(x² + 1 ) , f'(1) = ? 2,cho hs f(x) = e ngũ (2x + 1) , f'(1) = ? 23/07/2021 Bởi Hadley 1,cho hs f(x) = log2(x² + 1 ) , f'(1) = ? 2,cho hs f(x) = e ngũ (2x + 1) , f'(1) = ?
Đáp án: $\begin{array}{l}1)f\left( x \right) = \log 2\left( {{x^2} + 1} \right)\\ \Rightarrow f’\left( x \right) = \frac{{2.2x}}{{2\left( {{x^2} + 1} \right).\ln 10}} = \frac{{2x}}{{\left( {{x^2} + 1} \right).\ln 10}}\\ \Rightarrow f’\left( 1 \right) = \frac{1}{{\ln 10}} = \log e\\2)\\f\left( x \right) = {e^{2x + 1}}\\ \Rightarrow f’\left( x \right) = 2.{e^{2x + 1}}\\ \Rightarrow f’\left( 1 \right) = 2.{e^3}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
1)f\left( x \right) = \log 2\left( {{x^2} + 1} \right)\\
\Rightarrow f’\left( x \right) = \frac{{2.2x}}{{2\left( {{x^2} + 1} \right).\ln 10}} = \frac{{2x}}{{\left( {{x^2} + 1} \right).\ln 10}}\\
\Rightarrow f’\left( 1 \right) = \frac{1}{{\ln 10}} = \log e\\
2)\\
f\left( x \right) = {e^{2x + 1}}\\
\Rightarrow f’\left( x \right) = 2.{e^{2x + 1}}\\
\Rightarrow f’\left( 1 \right) = 2.{e^3}
\end{array}$