Calculus question
dy/dx(72x^2 - 4x^3)= dy/dx(72x^2)-dy/dx(4x^3)
=144x - 12x^2 => 12x(12-x)
If you don't want to open up the parenthesis then you can use the rule:
f*g' + g*f' = (fg)'
so:
f= x^2, f'=2x
g=72-4x, g'=-4
you get:
x^2*(-4)+(72-4x)(2x) = 12x(12-x)
Last edited by Red X; Mar 6, 2009 at 06:30 AM.
take the derivative of the individual pieces.
dy/dx(72x^2 - 4x^3)= dy/dx(72x^2)-dy/dx(4x^3)
=144x - 12x^2 => 12x(12-x)
If you don't want to open up the parenthesis then you can use the rule:
f*g' + g*f' = (fg)'
so:
f= x^2, f'=2x
g=72-4x, g'=-4
you get:
x^2*(-4)+(72-4x)(2x) = 12x(12-x)
dy/dx(72x^2 - 4x^3)= dy/dx(72x^2)-dy/dx(4x^3)
=144x - 12x^2 => 12x(12-x)
If you don't want to open up the parenthesis then you can use the rule:
f*g' + g*f' = (fg)'
so:
f= x^2, f'=2x
g=72-4x, g'=-4
you get:
x^2*(-4)+(72-4x)(2x) = 12x(12-x)
Funny how this makes sense to me
take the derivative of the individual pieces.
dy/dx(72x^2 - 4x^3)= dy/dx(72x^2)-dy/dx(4x^3)
=144x - 12x^2 => 12x(12-x)
If you don't want to open up the parenthesis then you can use the rule:
f*g' + g*f' = (fg)'
so:
f= x^2, f'=2x
g=72-4x, g'=-4
you get:
x^2*(-4)+(72-4x)(2x) = 12x(12-x)
dy/dx(72x^2 - 4x^3)= dy/dx(72x^2)-dy/dx(4x^3)
=144x - 12x^2 => 12x(12-x)
If you don't want to open up the parenthesis then you can use the rule:
f*g' + g*f' = (fg)'
so:
f= x^2, f'=2x
g=72-4x, g'=-4
you get:
x^2*(-4)+(72-4x)(2x) = 12x(12-x)
Im so glad I went through this in college and dont have to do it again!


