Đáp án:
$ 1) m<-\dfrac{4}{3}\\ 2)a)y'=\dfrac{2x^2-2x+1}{\sqrt{x^2+1}}\\ b)y'=\dfrac{4}{(x+3)^\tfrac{3}{2}\sqrt{3x+1}}$
Giải thích các bước giải:
$1) f(x)=\dfrac{1}{3}mx^3-\dfrac{m}{2}x^2+ (m+1)x -15\\ f'(x)=mx^2-mx+m+1\\ \circledast m=0, f'(x)=1 >0 (L)\\ \circledast m\ne 0, f'(x)<0 \, \forall \, x \in \mathbb{R}\\ \Leftrightarrow \left\{\begin{array}{l} \Delta <0\\m<0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m^2-4m(m+1)<0\\m<0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} −3m^2−4m<0\\m<0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} m>0\\m<-\dfrac{4}{3} \end{array} \right.\\m<0 \end{array} \right.\\ \Leftrightarrow m<-\dfrac{4}{3}\\ 2)\\ a)y=(x-2)\sqrt{x^2+1}\\ y'=(x-2)'\sqrt{x^2+1}+(x-2)\sqrt{x^2+1}'\\ =\sqrt{x^2+1}+(x-2)\dfrac{2x}{2\sqrt{x^2+1}}\\ =\sqrt{x^2+1}+(x-2)\dfrac{x}{\sqrt{x^2+1}}\\ =\dfrac{x^2+1+x(x-2)}{\sqrt{x^2+1}}\\ =\dfrac{2x^2-2x+1}{\sqrt{x^2+1}}\\ b)y=\sqrt{\dfrac{3x+1}{x+3}}\\ y'=\dfrac{\left(\dfrac{3x+1}{x+3}\right)'}{2\sqrt{\dfrac{3x+1}{x+3}}}\\ =\dfrac{\dfrac{8}{(x+3)^2}}{2\sqrt{\dfrac{3x+1}{x+3}}}\\ =\dfrac{4}{(x+3)^2\sqrt{\dfrac{3x+1}{x+3}}}\\ =\dfrac{4}{(x+3)^\tfrac{3}{2}\sqrt{3x+1}}$
