Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.y'=\frac{-sin\left( 6x^{3} +2\right)}{2\sqrt{cos^{2}\left( 3x^{3} +1\right)}} .9x^{2}\\ b.\ \ y=\frac{1}{9} x+\frac{20}{9} \ và\ \ \ y=\frac{1}{9} x+\frac{32}{9} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ y=\sqrt{cos^{2}\left( 3x^{3} +1\right)}\\ y'=\frac{\left[ cos^{2}\left( 3x^{3} +1\right)\right] '}{2\sqrt{cos^{2}\left( 3x^{3} +1\right)}}\\ =\frac{-2cos\left( 3x^{2} +1\right) .sin\left( 3x^{2} +1\right)}{2\sqrt{cos^{2}\left( 3x^{3} +1\right)}} .9x^{2}\\ =\frac{-sin\left( 6x^{3} +2\right)}{2\sqrt{cos^{2}\left( 3x^{3} +1\right)}} .9x^{2}\\ b.\ h'( x) =\frac{1}{( x-1)^{2}}\\ do\ tiếp\ tuyến\ song\ song\ với\ d\\ \Rightarrow hệ\ số\ góc\ k=\frac{1}{9}\\ \Rightarrow \frac{1}{( x-1)^{2}} =\frac{1}{9}\\ \Rightarrow ( x-1)^{2} =9\\ \Rightarrow x=4\ hoăc\ x=-2\\ TH1:\ x=4\Rightarrow y=\frac{8}{3}\\ PT\ tiếp\ tuyến:\ y=\frac{1}{9}( x-4) +\frac{8}{3}\\ hay\ y=\frac{1}{9} x+\frac{20}{9}\\ TH2:\ x=-2\Rightarrow y=\frac{10}{3}\\ PT\ tiếp\ tuyến:\ y=\frac{1}{9}( x+2) +\frac{10}{3}\\ hay\ y=\frac{1}{9} x+\frac{32}{9} \end{array}$