Đáp án:
$y'=\dfrac{2x+\dfrac{1}{\sqrt{x}}}{\cos^2(x^2+2\sqrt{x}+1)}\\$
Giải thích các bước giải:
$y=\tan(x^2+2\sqrt{x}+1)\\
\Rightarrow y'=\left [ \tan(x^2+2\sqrt{x}+1)\right ]'\\
=\dfrac{(x^2+2\sqrt{x}+1)'}{\cos^2(x^2+2\sqrt{x}+1)}\\
=\dfrac{(x^2)'+(2\sqrt{x})'+1'}{\cos^2(x^2+2\sqrt{x}+1)}\\
=\dfrac{2x+2\dfrac{x'}{2\sqrt{x}}}{\cos^2(x^2+2\sqrt{x}+1)}\\
=\dfrac{2x+\dfrac{1}{\sqrt{x}}}{\cos^2(x^2+2\sqrt{x}+1)}\\$
