\(f\left(x\right)=sin^3\left(ax\right)\)

=>\(f'\left(x\right)=3\cdot sin^2\left(ax\right)\cdot\left(sinax\right)'=3a\cdot sin^2\left(ax\right)\cdot cos\left(a\cdot x\right)\)

\(f'\left(\Omega\right)=3a\cdot sin^2\left(a\cdot\Omega\right)\cdot cos\left(a\cdot\Omega\right)\)

=>\(f'\left(\Omega\right)=3a\cdot0\cdot cos\left(a\cdot\Omega\right)=0\)

\(f\left(x\right)=\left(x+1\right)\cdot e^x\)

=>\(f'\left(x\right)=\left(x+1\right)'\cdot e^x+\left(x+1\right)\cdot\left(e^x\right)'\)

=>\(f'\left(x\right)=e^x+e^x\left(x+1\right)\)

\(f'\left(0\right)=e^0+e^0\left(0+1\right)=1+1=2\)

\(y=\left(x^2-2x+2\right)\cdot3^x\)

=>\(y'=\left(x^2-2x+2\right)'\cdot3^x+\left(x^2-2x+2\right)\cdot\left(3^x\right)'\)

=>\(y'=3^x\left(2x-2\right)+\left(x^2-2x+2\right)\cdot3^x\cdot ln3\)

\(y'=\dfrac{1'\cdot2^x-1\cdot\left(2^x\right)'}{2^{2x}}=\dfrac{-\left(2^x\right)'}{4^x}=\dfrac{-2^x\cdot ln2}{2^x\cdot2^x}=\dfrac{-ln2}{2^x}\)

Lời giải:

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

$\Rightarrow g'(x)=2xf(x)+x^2f'(x)$

$\Rightarrow g'(1)=2f(1)+f'(1)=2.1+1=3$

Lời giải:

$y'=((x^3+x)^{\frac{1}{2}})'=\frac{1}{2}(x^3+x)^{\frac{1}{2}-1}.(x^3+x)'$

\(=\frac{3x^2+1}{2\sqrt{x^3+x}}\)

\(y=\sqrt{3x^2+4}\)

=>\(y'=\dfrac{\left(3x^2+4\right)'}{2\sqrt{3x^2+4}}=\dfrac{6x}{2\sqrt{3x^2+4}}=\dfrac{3x}{\sqrt{3x^2+4}}\)

ĐKXĐ: \(4-log_2^2x>=0\)

=>\(log_2^2x< =4\)

=>\(-2< =log_2x< =2\)

=>\(-\dfrac{1}{4}< =x< =4\)

\(y=3^x+logx\)

=>\(y'=\left(3^x\right)'+\left(logx\right)'\)

=>\(y'=3^x\cdot ln3+\dfrac{1}{10\cdot lnx}\)

Chọn D