Đáp án:
$y'=\dfrac{6.(1+\sqrt{x})^{11} .x-\sqrt{x}(1+\sqrt{x}).6}{x^{7}\sqrt{x}}\\$
Giải thích các bước giải:
$y=\dfrac{(1+\sqrt{x})^{12}}{x^6}\\
\Rightarrow y'=\left ( \dfrac{(1+\sqrt{x})^{12}}{x^6} \right )'\\
=\dfrac{\left [(1+\sqrt{x})^{12} \right ]'x^6-(1+\sqrt{x})^{12}.(x^6)'}{(x^6)^2}\\
=\dfrac{12.(1+\sqrt{x})^{11} .(1+\sqrt{x})'.x^6-(1+\sqrt{x})^{12}.6x^5}{x^{12}}\\
=\dfrac{12.(1+\sqrt{x})^{11} .\dfrac{1}{2\sqrt{x}}x^6-(1+\sqrt{x})^{12}.6x^5}{x^{12}}\\
=\dfrac{12.(1+\sqrt{x})^{11} .x^6-2\sqrt{x}(1+\sqrt{x})^{12}.6x^5}{2x^{12}\sqrt{x}}\\
=\dfrac{6.(1+\sqrt{x})^{11} .x-\sqrt{x}(1+\sqrt{x})^{12}.6}{x^{7}\sqrt{x}}$
