What is the derivative of (x^4-1/(x^4+1)) ?

The derivative of (x^4-1/(x^4+1)) is (8x^3)/((x^4+1)^2)

What is the first derivative of (x^4-1/(x^4+1)) ?

The first derivative of (x^4-1/(x^4+1)) is (8x^3)/((x^4+1)^2)

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

\frac{8 x ^{3}}{( x ^{4} + 1 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

4 x^{3} (x^{4} + 1) - 4 x^{3} (x^{4} - 1) : 8 x^{3}

= \frac{8 x ^{3}}{( x ^{4} + 1 ) ^{2}}

\frac{d}{d x} ((x^{4} - 3 x^{2} + 5)^{3})

n = 1 \sum \infty \frac{n}{( 2 n ) !}

f (x) = x^{2} - 8 x - 8

\frac{\partial}{\partial x} (x^{6} y^{8} - x^{5} y^{7})

x \to 0 + lim (tan ((3 x)^{x}))

What is the derivative of (x^4-1/(x^4+1)) ?
The derivative of (x^4-1/(x^4+1)) is (8x^3)/((x^4+1)^2)
What is the first derivative of (x^4-1/(x^4+1)) ?
The first derivative of (x^4-1/(x^4+1)) is (8x^3)/((x^4+1)^2)