It is #10*e^x*e^(2x)*e^(3x)*e^(4x)#

#f(x)=e^x*e^(2x)*e^(3x)*e^(4x)=e^(x+2x+3x+4x)=e^(10x)#

so #f'(x)=3^(10x)*10=10e^(10x)# You could leave it that way, but a teacher who would ask such a question I would give the more cumbersome answer stated above.

(note that since we got #f(x)=e^(10x)#, we have now found that #f'(x)=10f(x)# -- that's why we can quickly rewrite as I did in the first line.)

**(Or worse yet, I'd use the multiple factor product rule:**

#d/(dx)(f_1f_2f_3 * * * f_n)=f_1'f_2f_3 * * * f_n+f_1f_2'f_3 * * * f_n+f_1f_2f_3' * * * f_n+* * * +f_1f_2f_3 * * * f_n'#

So for this function we could write:

#f'(x)=color(red)(e^x)*e^(2x)*e^(3x)*e^(4x) + e^x*color(red)(e^(2x)*2)*e^(3x)*e^(4x)+e^x*e^(2x)*color(red)(e^(3x)*3)*e^(4x)+e^x*e^(2x)*e^(3x)*color(red)(e^(4x)*4)#

(derivatives are in #color(red)(red)#)

(It's correct, but I'd probably lose points for not simplifying.)