What is the derivative of sqrt((x^2-x^4/8)) ?

The derivative of sqrt((x^2-x^4/8)) is (-2x^2+1)/(2sqrt(2)\sqrt{-x^2+1)}

What is the first derivative of sqrt((x^2-x^4/8)) ?

The first derivative of sqrt((x^2-x^4/8)) is (-2x^2+1)/(2sqrt(2)\sqrt{-x^2+1)}

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Popular >

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

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

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\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}

\frac{d}{d x} (x - arcsin (x))

y^{^{'}} = \frac{sin ( x )}{cos ( y )}

f (x) = 4 lo g_{2} (x)

\int \frac{7}{x ^{2} 1 + x ^{2}} d x

x^{2} e^{- x^{2}}

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