What is the derivative of f(x)=(-8x)/(x^2+1) ?

The derivative of f(x)=(-8x)/(x^2+1) is -(8(-x^2+1))/((x^2+1)^2)

What is the first derivative of f(x)=(-8x)/(x^2+1) ?

The first derivative of f(x)=(-8x)/(x^2+1) is -(8(-x^2+1))/((x^2+1)^2)

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derivative of

f (x) = \frac{- 8 x}{x ^{2} + 1}

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

Solution steps

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

Simplify

\frac{- 8 x}{x ^{2} + 1} : - \frac{8 x}{x ^{2} + 1}

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

Take the constant out:

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

= - 8 \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}}

= - 8 \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

= - 8 \cdot \frac{1 \cdot ( x ^{2} + 1 ) - 2 xx}{( x ^{2} + 1 ) ^{2}}

Simplify

- 8 \cdot \frac{1 \cdot ( x ^{2} + 1 ) - 2 xx}{( x ^{2} + 1 ) ^{2}} : - \frac{8 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

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

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

(x + x y^{2}) d x + e^{x^{2}} y d y = 0

\int_{- 2}^{x^{2} - 3 x} e^{t^{2}} d t

y = x^{\frac{4}{5}} (x - 3)

\frac{d}{d x} (\frac{- 2 x ^{2} + 25}{25 - x ^{2}})

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