What is the derivative of-(6x/(1-3x^2)) ?

The derivative of-(6x/(1-3x^2)) is -(6(1+3x^2))/((1-3x^2)^2)

What is the first derivative of-(6x/(1-3x^2)) ?

The first derivative of-(6x/(1-3x^2)) is -(6(1+3x^2))/((1-3x^2)^2)

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\frac{d}{d x} (- \frac{6 x}{1 - 3 x ^{2}})

- \frac{6 ( 1 + 3 x ^{2} )}{( 1 - 3 x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (- \frac{6 x}{1 - 3 x ^{2}})

Take the constant out:

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

= - 6 \frac{d}{d x} (\frac{x}{1 - 3 x ^{2}})

Apply the Quotient Rule:

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

= - 6 \cdot \frac{\frac{d x}{d x} ( 1 - 3 x ^{2} ) - \frac{d}{d x} ( 1 - 3 x ^{2} ) x}{( 1 - 3 x ^{2} ) ^{2}}

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

\frac{d}{d x} (1 - 3 x^{2}) = - 6 x

= - 6 \cdot \frac{1 \cdot ( 1 - 3 x ^{2} ) - ( - 6 x ) x}{( 1 - 3 x ^{2} ) ^{2}}

Simplify

- 6 \cdot \frac{1 \cdot ( 1 - 3 x ^{2} ) - ( - 6 x ) x}{( 1 - 3 x ^{2} ) ^{2}} : - \frac{6 ( 1 + 3 x ^{2} )}{( 1 - 3 x ^{2} ) ^{2}}

= - \frac{6 ( 1 + 3 x ^{2} )}{( 1 - 3 x ^{2} ) ^{2}}

\int (1 - cos^{2} (x)) d x

\int sin (2 y) d y

x \to 2 lim (x + \frac{1}{x - 2})

y = 4^{\frac{x}{l n ( 2 )}}, at x = ln (16)

\int \frac{1}{5 x ^{2}} d x

What is the derivative of-(6x/(1-3x^2)) ?
The derivative of-(6x/(1-3x^2)) is -(6(1+3x^2))/((1-3x^2)^2)
What is the first derivative of-(6x/(1-3x^2)) ?
The first derivative of-(6x/(1-3x^2)) is -(6(1+3x^2))/((1-3x^2)^2)