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Find f"(x) for each function. Then find f"(0) and f"(2) f(x)=-x/1-x^2
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\(\frac{-x}{1-x^2}\)?
Yes
do you know how to find the first derivative?
no
i missed the lesson about this
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\((\frac{f(x)}{g(x)})'=\frac{f'(x)g(x)-g'(x)f(x)}{(g(x))^2}\)
so you have \(f(x) = -x, g(x) = 1-x^2\)
okay
\(f'(x)=-1,g'(x)=-2x \)so \((\frac{-x}{1-x^2})'=\frac{-1(1-x^2)-(-2x)(-x)}{(1-x^2)^2}\)now simplify
how did 1-x^2 turned to -2x
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i dont know how to simplify this
you need to go back and review, it seems you are not at the level you shuold be to do these problems if f(x) = 1-x^2 then f'(x) = -2x
okay now simplifying this problem will be (-1+1x^2)+(2x+1)/1-x^4
am i correct
@jdoe0001
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@@AccessDenied @eyust707
I need help simplifying this
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