EM THAM KHẢO :

12.

$y = (x^2 + 1)^{\sin x}$

$\Rightarrow y = e^{\sin x \ln(x^2 + 1)}$

$\Rightarrow y' = \left(e^{\sin x \ln(x^2 + 1)}\right)'$

$= e^{\sin x \ln(x^2 + 1)} \cdot (\sin x \ln(x^2 + 1))'$

$= (x^2 + 1)^{\sin x} \cdot \left( (\sin x)' \ln(x^2 + 1) + \sin x (\ln(x^2 + 1))' \right)$

$= (x^2 + 1)^{\sin x} \cdot \left( \cos x \ln(x^2 + 1) + \sin x \cdot \dfrac{(x^2 + 1)'}{x^2 + 1} \right)$

$= (x^2 + 1)^{\sin x} \cdot \left( \cos x \ln(x^2 + 1) + \sin x \cdot \dfrac{2x}{x^2 + 1} \right)$

$= (x^2 + 1)^{\sin x} \left( \cos x \ln(x^2 + 1) + \dfrac{2x \sin x}{x^2 + 1} \right)$