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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\int \frac{8 - 128 x}{9 - 64 x ^{2}} d x

\frac{\partial}{\partial t} (t cos (s))

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

(arccos (x))^{^{'}}

\frac{d ^{4}}{d x ^{4}} (5 e^{x^{2}})

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