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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derivative of

- \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}}

\frac{d}{d x} (ln ((arcsin (x))^{2}))

\frac{140000}{x}

\int \frac{1}{x ^{2} - 3 x - 15} d x

\int 40 + 6 x - x^{2} d x

\int \frac{1}{x ( I n x ) ^{2}} d 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)