Evaluate d/dx

^{3}√(ln(7-x^2)) at x=1^{ }Evaluate d/dx ^{3}√(ln(7-x^2)) at x=1^{
}

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f(X) = (ln(7-X^{2}))^{(1/3)}

f'(X) = (1/3)* (ln(7-X2))^{(-2/3)} * (1/(7-X^{2})) * (-2X)

for X=1 we have:

f'(X) = 0.2259

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## Comments

^{2}-7)log2/3(7-x^{2}). How would i go about solving for x=1? do i just plug in 1 where ever there is an x?