The value of the derivative of:

y= 0.5x^3+2x^2+3x+1500
at x=5 is.....

Question

The value of the derivative of:

y= 0.5x^3+2x^2+3x+1500
at x=5 is.....

anonymous · Answer

y' = 1.5x^2+4x+3
y'(3) = 1.5(3)^2+4(3)+3
y'(5) = 45+20+3 = 68

I think I did that math right. Please double check it because I did it in my head.

anonymous · Answer

put x=5 in 1.5x^2 + 4x + 3

anonymous · Answer

I reposted. I still have 3's in there. But the math is done right now. Sorry

anonymous · Answer

thanks for helping me..^^

anonymous · Answer

You're welcome. Do you understand how to differentiate?

anonymous · Answer

and the answer would be 60.5.
@teic85 i did not understand your calculations :O
1.5*(5^2)= 45?

anonymous · Answer

do you believe 1.5*9 gives 45?

anonymous · Answer

lol sorry, i'm tired and im not using paper and pencil

anonymous · Answer

you did the whole answer wrong!!!!
put x=5 in the equation!!

anonymous · Answer

Once you have y', you can just plug in the 5.

anonymous · Answer

I'm 99% my y' is correct.

anonymous · Answer

$$y' = 1.5x^2+4x+3  $$$$y'(5) = 1.5(5)^2+4(5)+3 $$ $$y'(5) = 37.5+20+3 = 62.5 $$