- g(x) = (x+1/x)
g'(x) = 1 - 1/x²
f(x) = -ln(g(x)) = - g'(x)/g(x) = - (1-1/x²)/(x+1/x)
= (1-x²)/(x + x³)
Why ask the same question so many times?00
- I guess (?) that's a minus sign just after the 2nd t ?
d[ -log(x + 1/x) ]/dx
= -[1/(x + 1/x)]*(1 - 1/x^2)
= -[x/(x^2 + 1)]*[(x^2-1)/x^2]
= (1 - x^2)/(x^3 + x).00
- I wonder if the fact that as a child I had problems with multiplication, maybe some relationship with my Asperger Syndrome?
- Help solve?
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