0 bình luận về “Tính đạo hàm của hàm số
y= x ² √x (sinx)”
Đáp án:
$\begin{array}{l} y = {x^2}\sqrt x .{\mathop{\rm s}\nolimits} {\rm{inx}} = {x^{\frac{5}{2}}}.\sin x\\ \Rightarrow y’ = \frac{5}{2}.{x^{\frac{3}{2}}}.\sin x + {x^{\frac{5}{2}}}.\cos x\\ = \frac{5}{2}x\sqrt x .\sin x + {x^2}\sqrt x .\cos x \end{array}$
Đáp án:
$\begin{array}{l}
y = {x^2}\sqrt x .{\mathop{\rm s}\nolimits} {\rm{inx}} = {x^{\frac{5}{2}}}.\sin x\\
\Rightarrow y’ = \frac{5}{2}.{x^{\frac{3}{2}}}.\sin x + {x^{\frac{5}{2}}}.\cos x\\
= \frac{5}{2}x\sqrt x .\sin x + {x^2}\sqrt x .\cos x
\end{array}$
$y=x^2.\sqrt{x\sin x}$
$y’=(x^2)’\sqrt{x\sin x}+x^2(\sqrt{x\sin x})’$
$=2x\sqrt{x\sin x}+x^2.(\sqrt{x})’.\sqrt{\sin x}+x^2\sqrt{x}(\sqrt{\sin x})’$
$=2x\sqrt{x\sin x}+\dfrac{x^2}{2\sqrt{x}}\sqrt{\sin x}+x^2\sqrt{x}.\dfrac{(\sin x)’}{2\sqrt{\sin x}}$
$=2x\sqrt{x\sin x}+\dfrac{x\sqrt{x}}{2}\sqrt{\sin x}+x^2\sqrt{x}\dfrac{\cos x}{2\sqrt{\sin x}}$