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bio_man bio_man
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Posts: 33318
6 years ago


#1. Start with the first derivative:

\(\frac{dy}{dx}=\frac{2}{3x^2}\)

Now the second derivative:

\(\frac{d^2y}{dx}=\frac{-4}{3x^3}\)



#2. We have to take the derivative with respect to a

\(y\ =\ x(a^2-x^2)^{\left(\frac{1}{2}\right)}\)

\(\frac{dy}{dx}=\ \left(\frac{1}{2}\right)x(a^2-x^2)^{\left(-\frac{1}{2}\right)}\cdot \left(2a-x^2\right)\)

\(\frac{dy}{dx}=\ \left(\frac{x(a^2-x^2)^{\left(-\frac{1}{2}\right)}\cdot \left(2a-x^2\right)}{2}\right)\)

Simplify more:

\(\frac{dy}{dx}=\ \left(\frac{x\cdot \left(2a-x^2\right)}{2(a^2-x^2)^{\left(\frac{1}{2}\right)}}\right)\)

You can change to radical form if you like:

\(\frac{dy}{dx}=\ \left(\frac{x\left(2a-x^2\right)}{2\sqrt{a^2-x^2}}\right)\)
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wrote...
5 years ago
Great
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