What is the derivative of (x^3+x/(x^4-2)) ?

The derivative of (x^3+x/(x^4-2)) is (-x^6-3x^4-6x^2-2)/((x^4-2)^2)

What is the first derivative of (x^3+x/(x^4-2)) ?

The first derivative of (x^3+x/(x^4-2)) is (-x^6-3x^4-6x^2-2)/((x^4-2)^2)

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\frac{d}{d x} (\frac{x ^{3} + x}{x ^{4} - 2})

\frac{- x ^{6} - 3 x ^{4} - 6 x ^{2} - 2}{( x ^{4} - 2 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{3} + x}{x ^{4} - 2})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (x^{3} + x) = 3 x^{2} + 1

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

= \frac{( 3 x ^{2} + 1 ) ( x ^{4} - 2 ) - 4 x ^{3} ( x ^{3} + x )}{( x ^{4} - 2 ) ^{2}}

Expand

(3 x^{2} + 1) (x^{4} - 2) - 4 x^{3} (x^{3} + x) : - x^{6} - 3 x^{4} - 6 x^{2} - 2

= \frac{- x ^{6} - 3 x ^{4} - 6 x ^{2} - 2}{( x ^{4} - 2 ) ^{2}}

y = 2 e^{- x} + e^{3 x}

n = 1 \sum \infty \frac{n}{4 n ^{3} + 1}

h (x) = sec^{6} (x)

\int \frac{e ^{x}}{1 - 25 e ^{2 x}} d x

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{2}})

What is the derivative of (x^3+x/(x^4-2)) ?
The derivative of (x^3+x/(x^4-2)) is (-x^6-3x^4-6x^2-2)/((x^4-2)^2)
What is the first derivative of (x^3+x/(x^4-2)) ?
The first derivative of (x^3+x/(x^4-2)) is (-x^6-3x^4-6x^2-2)/((x^4-2)^2)