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Popular >

(sqrt((x^2-x^4)/8))^'

Solution

(\frac{x ^{2} - x ^{4}}{8})^{^{'}}

Solution

\frac{- 2 x ^{2} + 1}{2 2 - x ^{2} + 1}

Solution steps

(\frac{x ^{2} - x ^{4}}{8})^{'}

Apply the chain rule:

\frac{1}{2 \frac{x ^{2} - x ^{4}}{8}} (\frac{x ^{2} - x ^{4}}{8})^{'}

= \frac{1}{2 \frac{x ^{2} - x ^{4}}{8}} (\frac{x ^{2} - x ^{4}}{8})^{'}

(\frac{x ^{2} - x ^{4}}{8})^{'} = \frac{x ( - 2 x ^{2} + 1 )}{4}

= \frac{1}{2 \frac{x ^{2} - x ^{4}}{8}} \cdot \frac{x ( - 2 x ^{2} + 1 )}{4}

Simplify

\frac{1}{2 \frac{x ^{2} - x ^{4}}{8}} \cdot \frac{x ( - 2 x ^{2} + 1 )}{4} : \frac{- 2 x ^{2} + 1}{2 2 - x ^{2} + 1}

= \frac{- 2 x ^{2} + 1}{2 2 - x ^{2} + 1}

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Frequently Asked Questions (FAQ)

What is the (sqrt((x^2-x^4)/8))^' ?
The (sqrt((x^2-x^4)/8))^' is (-2x^2+1)/(2sqrt(2)\sqrt{-x^2+1)}
What is the first (sqrt((x^2-x^4)/8))^' ?
The first (sqrt((x^2-x^4)/8))^' is (-2x^2+1)/(2sqrt(2)\sqrt{-x^2+1)}