Differentiation

[Guest]

Y= 2x^3/3 (2X to the power of 3 divided by 3)

DY/DX= 2x^2 (2x to the power of 2)

how? does this happen HELP!

[Guest]

[tex]\frac{2x^3}{3}=\frac{2}{3}x^3 = \frac{2}{3}*3*x^2 = 2x^2[/tex]

[Guest]

[tex]\frac{2x^3}{3}=\frac{2}{3}x^3 = \frac{2}{3}*3*x^2 = 2x^2[/tex]

thanks i figured it out but thanks anyway

just wondering however what you did mathmatically to go from

[tex]\frac{2x^3}{3}[/tex]

to

[tex]\frac{2}{3}x^3[/tex]

i have another that is

Y=-[tex]\frac{5}{2e^3^x}[/tex]

please note that when Y=e^kx ......DY/DX= ke^kx

and to get the answer the same thing is done...so help again please ledgend!