Let f(x)=det(x2excos(x)ln(x+1))f(x) = \det \begin{pmatrix} x^2 & e^x \\ \cos(x) & \ln(x+1) \end{pmatrix}f(x)=det(x2cos(x)exln(x+1)). Find f′(0)f'(0)f′(0).
f′(0)=0f'(0) = 0f′(0)=0
f′(0)=1f'(0) = 1f′(0)=1
f′(0)=−1f'(0) = -1f′(0)=−1
f′(0)=ef'(0) = ef′(0)=e