a)$y=x^2e^{x^3}cosx=>ln|y|=2ln|x|+x^3+ln|cosx|$ Lấy đạo hàm hai vế ta được: $\frac{y'}{y}=\frac{2}{x}+3x^2-tgx$ $=>y'=y(\frac{2}{x}+3x^2-cotgx)=x.e^{x^3}.cosx.(2+3x^3-tgx)$ b)$lim_{x->0}y'=lim_{x->0}x.e^{x^3}.cosx.(2+3x^3-tgx)=lim_{x->0}0.e^{0^3}.cos0.(2+3.0^3-tg0)=0$
a)$y=x^2e^{x^3}cosx=>ln|y|=2ln|x|+x^3+ln|cosx|$ Lấy đạo hàm hai vế ta được: $\frac{y'}{y}=\frac{2}{x}+3x^2-tgx$ $=>y'=y(\frac{2}{x}+3x^2-cotgx)=x.e^{x^3}.cosx.(2+3x^3-tgx)$ b)$lim_{x->0}y'=lim_{x->0}x.e^{x^3}.cosx.(2+3x^3-tgx)=lim_{x->0}0.e^{0^3}.cos0.(2+3.0^3-tg0)=0$
