GIUP BAN NGUYEN VAN AN

Cho hàm số f  2 Tính f  x có đạo hàm � thỏa  x   f  x    x  1 f �  x   ex Giải: ( x)  e x  x   f  x    x  1 f �  x   e x � ( x  1) f ( x)  f ( x)  ( x  1) f � � (x + 1)f (x) + (x + 1)� f (x) + (x + 1)f � (x) = ex � x � (x + 1)f (x) + � ( x + 1) f (x)� � �= e (*) Nhân vế (*) cho ex ta �x � (x + 1)f (x)� ex + � e = e2x ( x + 1) f (x)� � � � � �x 2x �� (x + 1)f (x)� (ex )� +� ( x + 1) f (x)� � �e = e � � � �� (x + 1)f (x).ex �= e2x � � 2 2 � � �� (x + 1)f (x)e �dx = � e2xdx � (x + 1)f (x)ex = e2x � � 0 x � (x + 1)f (x) = ex 1 � 3ff(2) - (0) = e2 2 � f (2) = e2 f  0 