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Mathematics 20 Online
OpenStudy (anonymous):

d^2y/dx^2 = (dy dx)^2 true or false

OpenStudy (primeralph):

You tell me.

OpenStudy (anonymous):

i think its false

OpenStudy (primeralph):

Show me why.

OpenStudy (anonymous):

because that would imply that the second deriv is equal to the first deriv squared and that is not always the case

OpenStudy (anonymous):

does that make sense?

OpenStudy (primeralph):

Show me a mathematical illustration with an example to support your claim.

OpenStudy (anonymous):

like for example take 2x^3-1 the deriv is 6x^2 the second derivative 12x and 12x does not equal (6x^2)^2

OpenStudy (primeralph):

Good job.

OpenStudy (anonymous):

thank you :)

OpenStudy (primeralph):

You did all the work. Didn't need my help.

OpenStudy (anonymous):

that wasnt the problem tho im having trouble with an absolute value one

OpenStudy (anonymous):

d/dx |x2 + x| = |2x + 1|

OpenStudy (anonymous):

now.... if i am thinking on the right track the left side of the equation would be x^2 +x over the absolute value of the function times the deriv which is 2x

OpenStudy (anonymous):

but i dont know how to calculate the problem.. so i might not even be on the right track

OpenStudy (primeralph):

Not really.

OpenStudy (primeralph):

You have to set the conditions for the existence of the abs function. You need to make it piecewise.

OpenStudy (primeralph):

|dw:1381978278075:dw|

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