What is the derivative of-(6x/(x^2+1)) ?

The derivative of-(6x/(x^2+1)) is -(6(-x^2+1))/((x^2+1)^2)

What is the first derivative of-(6x/(x^2+1)) ?

The first derivative of-(6x/(x^2+1)) is -(6(-x^2+1))/((x^2+1)^2)

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\frac{d}{d x} (- \frac{6 x}{x ^{2} + 1})

- \frac{6 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (- \frac{6 x}{x ^{2} + 1})

Take the constant out:

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

= - 6 \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}}

= - 6 \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

= - 6 \cdot \frac{1 \cdot ( x ^{2} + 1 ) - 2 xx}{( x ^{2} + 1 ) ^{2}}

Simplify

- 6 \cdot \frac{1 \cdot ( x ^{2} + 1 ) - 2 xx}{( x ^{2} + 1 ) ^{2}} : - \frac{6 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

= - \frac{6 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

\int 9 - 4 x d x

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

x \to - 1 lim (x^{3} - x^{2} - 5 x + 3)

r = 3 + 2 sin (θ)

y = x x

What is the derivative of-(6x/(x^2+1)) ?
The derivative of-(6x/(x^2+1)) is -(6(-x^2+1))/((x^2+1)^2)
What is the first derivative of-(6x/(x^2+1)) ?
The first derivative of-(6x/(x^2+1)) is -(6(-x^2+1))/((x^2+1)^2)