What is the derivative of (2x)/(x^4+1) ?

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

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

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

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derivative of

\frac{2 x}{x ^{4} + 1}

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

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

x \to 0 + lim ((1 + 4 x)^{\frac{5}{x}})

\frac{d y}{d x}, y^{5} = x^{7}

x^{2} + 3 x + \frac{1}{x ^{2}}

\int \frac{arcsin ( 4 x )}{1 - 16 x ^{2}} d x

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