What's my problem with this second derivative question? First derivative: f'(x)= 20x / (x^2 - 4)^2 Second derivative correct answer: f"(x)= -20(3x^2 + 4) / (x^2 - 4)^3 So I used the quotient rule to find the second derivative, and I ended up with f"(x)= 20(x^2 - 4)^2 - 80x(x^2 - 4) / (x^2 - 4)^4. Did I do something wrong or do I need to take further steps to get the final answer?
\(\large f'(x) = \frac{20x}{(x^2-4)^2}\) \(\large f''(x) = \frac{20(x^2-4)^2 - 80x(x^2-4)}{(x^2-4)^4}\)
your second derivative is correct ! just you need to simplify it further
how?
factor \((x^2-4)\) in the numerator, and cancel it wid denominator
\(\large f'(x) = \frac{20x}{(x^2-4)^2}\) \(\large f''(x) = \frac{20(x^2-4)^2 - 80x(x^2-4)}{(x^2-4)^4}\) \(\large f''(x) = \frac{(x^2-4)[20(x^2-4) - 80x]}{(x^2-4)^4}\)
Okay... but that still doesn't end up with the correct answer :(...
Even after canceling out (x^2-4) from both the numerator and the denominator
\(\large f'(x) = \frac{20x}{(x^2-4)^2}\) \(\large f''(x) = \frac{20(x^2-4)^2 - 80x(x^2-4)}{(x^2-4)^4}\) \(\large f''(x) = \frac{(x^2-4)[20(x^2-4) - 80x]}{(x^2-4)^4}\) \(\large f''(x) = \frac{20(x^2-4) - 80x}{(x^2-4)^3}\)
The book says the correct final answer is f"(x)= -20(3x^2 + 4) / (x^2 - 4)^3...
found the mistake !
\(\large f'(x) = \frac{20x}{(x^2-4)^2}\) \(\large f''(x) = \frac{20(x^2-4)^2 - 20x*2x*2*(x^2-4)}{(x^2-4)^4}\) \(\large f''(x) = \frac{20(x^2-4)^2 - 80x^2(x^2-4)}{(x^2-4)^4}\) \(\large f''(x) = \frac{(x^2-4)[20(x^2-4) - 80x^2]}{(x^2-4)^4}\) \(\large f''(x) = \frac{20(x^2-4) - 80x^2}{(x^2-4)^3}\)
now simplify
\[\frac{ 20 (x^2-4) - 80x^2 }{ (x^2 - 4)^3} \] \[\frac{ 20x^2 -80 - 80x^2 }{ (x^2 - 4)^3} \] \[\frac{ -60x^2 - 80 }{ (x^2 - 4)^3} \] \[\frac{ -20 (3x^2 + 4) }{ (x^2 - 4)^3} \]
I finally got it!! Thank you so much!
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