Can anyone solve this one? I got stuck.. f(x) = 4/ (x^2 + 8) I need f''(x) second derivative for this
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Originally Posted by dkssudgktpdy Can anyone solve this one? I got stuck.. f(x) = 4/ (x^2 + 8) I need f''(x) second derivative for this Can you get the first derivative? Please show all your work.
Originally Posted by mr fantastic Can you get the first derivative? Please show all your work. yup. f(x) = 4 (x^2 +8)^-1 = -4 (x^2 +8)^-2 (2x) = -8x (x^2 +8)^-2 f''(x)= 16x (x^2 +8)^-3 (2x) = 32x (x^2 +8)^-3 0 = 32x/ (x^2 +8)^3 x=0 and.. 0 is not the right answer. Where did I make a mistake?
Originally Posted by dkssudgktpdy yup. f(x) = 4 (x^2 +8)^-1 = -4 (x^2 +8)^-2 (2x) = -8x (x^2 +8)^-2 f''(x)= 16x (x^2 +8)^-3 (2x) = 32x (x^2 +8)^-3 0 = 32x/ (x^2 +8)^3 x=0 and.. 0 is not the right answer. Where did I make a mistake? You need to use the product rule or quotient. $\displaystyle \frac{d^2y}{dx^2}=\frac{(x^2+8)(-8)-(-8x)(2(x^2+8)(2x))}{(x^2+8)^4}$
Originally Posted by TheEmptySet You need to use the product rule or quotient. $\displaystyle \frac{d^2y}{dx^2}=\frac{(x^2+8)(-8)-(-8x)(2(x^2+8)(2x))}{(x^2+8)^4}$ I got it. Thank you!
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