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About LinkBacksI have the following:
Where
My question is, does?
Try keeping up to date with what happens by reading the posts here: http://www.mathhelpforum.com/math-help/f26/ ! (I have applied the temporary fix to your latex).Originally Posted by Bwts
I have the following:
Where
My question is, does
ps I have no idea why my latex code isnt working on your site?
It does not give you the same answer, if you differentiate with respect to a variable and with respect to that variable squared.
For example, consider the function f(x)=x^2. If you take the derivative of this function with respect to x, you of course get 2x. If you take the derivative with respect to x^2 (which is df/dx^2), then you get the answer 1. Why? In df/dx^2, you treat x^2 as a single variable, so you might as well call it something less confusing, say v=x^2. Then the problem of finding df/dx^2, when f(x)=x^2, translates to finding df/dv, when f(v) = v, which is 1.
Similarly, g(x)=x^4 is 4x^3 upon differentiation with respect to x, but with respect to x^2, it becomes 2x^2 (since x^4 = (x^2)^2, and (x^2)^2 differentiated with respect to x^2 is 2x^2).
I'm thinking that maybe you have an easier time relating to Fantastic's suggestion.
By the chain rule, we have that
df/dx = df/dx^2 * dx^2/dx,
but since dx^2/dx = x^2, we have that df/dx^2 =(df/dx)/(2x). So to find the derivative with respect to x^2, you find the derivative of f with respect to x, and then divides this answer by 2x.
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