Tìm đạo hàm cấp hai mỗi hàm số sau:
a) \(f\left( x \right) = \frac{1}{{3x + 5}};\)
b) \(g\left( x \right) = {2^{x + 3{x^2}}}.\)
Tính \(f’\left( x \right)\) rồi tính \(f”\left( x \right).\)
a) \(f\left( x \right) = \frac{1}{{3x + 5}} \Rightarrow f’\left( x \right) = - \frac{3}{{{{\left( {3x + 5} \right)}^2}}} \Rightarrow f”\left( x \right) = - 3.\frac{{ - 2\left( {3x + 5} \right).3}}{{{{\left( {3x + 5} \right)}^4}}} = \frac{{18}}{{{{\left( {3x + 5} \right)}^3}}}.\)
b) \(g\left( x \right) = {2^{x + 3{x^2}}} \Rightarrow g’\left( x \right) = \left( {6x + 1} \right){2^{x + 3{x^2}}}\ln 2\)
\( \Rightarrow f”\left( x \right) = \ln 2.\left[ {{{6.2}^{x + 3{x^2}}} + \left( {6x + 1} \right).\left( {6x + 1} \right){2^{x + 3{x^2}}}\ln 2} \right] = \ln {2.2^{x + 3{x^2}}}\left[ {6 + {{\left( {6x + 1} \right)}^2}\ln 2} \right].\)