quick calculus question...

[50]

So I had a question on a test to find the second derviative of (x*d/dx(2x-3x^3)

this is a really simple question, but since u have an x next to d/dx wouldn't that just be dx/dx? Thanks

 

[akubi]

uh no

 

[TheoPetro]

x d/dx would be one, the derivative of x with respect to x is one.

 

[Heisenberg]

x * (d/dx)(2x-3x^3) = x * (2-9x^2) = 2x-9x^3, then the derivative of that is 2-27x^2 is the way I read it.

Edit: I can't mulitply.

 

[akubi]

Originally posted by: Heisenberg
x * (d/dx)(2x-3x^3) = x * (2-9x^2) = 2x-18x^2 is the way I read it.

it is the only way to read it.

edit: um except you make a mistake in the last step...

 

[50]

So you can't remove an x from the d/dx?

 

[CycloWizard]

No. d/dx operates on whatever is inside the parentheses to its right. The x in front is multiplied by the derivative, not actually being operated on by the derivative d/dx. In this case, you would take the derivative of 2x-3x^3. This could be rewritten as x*[d(2x)/dx-d(3x^3)/dx], with the result being x(2-9x^2)=2x-9x^3.

 

[Born2bwire]

Originally posted by: 50
So I had a question on a test to find the second derviative of (x*d/dx(2x-3x^3)

this is a really simple question, but since u have an x next to d/dx wouldn't that just be dx/dx? Thanks

In addition, the actual operation on the product can be found by using the chain rule, that is, d/dx(x*d/dx(2x-3x^3)) = d/dx(2x-3x^3)+x*d^2/dx^2(3x-3x^3)