Bài tập 33 trang 78 SBT Toán 11 Tập 2 Cánh diều - CD

a) Xét hàm số $f\left(x\right)=\frac{1}{3x+5}.$

Ta có: $f^{\text{'}}\left(x\right)=-\frac{{\left(3x+5\right)}^{\text{'}}}{{\left(3x+5\right)}^2}=\frac{-3}{{\left(3x+5\right)}^2};$

$f^{\text{'}\text{'}}\left(x\right)=\left(-3\right)\cdot \frac{-{\left[{\left(3x+5\right)}^2\right]}^{\text{'}}}{{\left[{\left(3x+5\right)}^2\right]}^2}=\frac{3\cdot 2\left(3x+5\right)\cdot 3}{{\left(3x+5\right)}^4}=\frac{18}{{\left(3x+5\right)}^3}.$

b) Xét hàm số $g\left(x\right)=2^{x+3x^2}.$ Ta có:

Ta có: $g^{\text{'}}\left(x\right)={\left(x+3x^2\right)}^{\text{'}}\text{ln}2\cdot 2^{x+3x^2}=\left(6x+1\right)\text{ln}2\cdot 2^{x+3x^2}.$

${g^{\text{'}}}^{\text{'}}\left(x\right)=\text{ln}2\left[{\left(6x+1\right)}^{\text{'}}\cdot 2^{x+3x^2}+\left(6x+1\right)\cdot {\left(2^{x+3x^2}\right)}^{\text{'}}\right]$

$=\text{ln}2\left[6\cdot 2^{x+3x^2}+\left(6x+1\right)\cdot \left(6x+1\right)\text{ln}2\cdot 2^{x+3x^2}\right]$

$=6\text{ln}2\cdot 2^{x+3x^2}+{\left[\left(6x+1\right)\text{ln}2\right]}^2\cdot 2^{x+3x^2}.$

-- Mod Toán 11 HỌC247