What is the derivative of (x^2-5)/(1-3x^2) ?

The derivative of (x^2-5)/(1-3x^2) is -(28x)/((1-3x^2)^2)

What is the first derivative of (x^2-5)/(1-3x^2) ?

The first derivative of (x^2-5)/(1-3x^2) is -(28x)/((1-3x^2)^2)

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derivative of

\frac{x ^{2} - 5}{1 - 3 x ^{2}}

- \frac{28 x}{( 1 - 3 x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 5}{1 - 3 x ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} - 5 ) ( 1 - 3 x ^{2} ) - \frac{d}{d x} ( 1 - 3 x ^{2} ) ( x ^{2} - 5 )}{( 1 - 3 x ^{2} ) ^{2}}

\frac{d}{d x} (x^{2} - 5) = 2 x

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

= \frac{2 x ( 1 - 3 x ^{2} ) - ( - 6 x ) ( x ^{2} - 5 )}{( 1 - 3 x ^{2} ) ^{2}}

Simplify

\frac{2 x ( 1 - 3 x ^{2} ) - ( - 6 x ) ( x ^{2} - 5 )}{( 1 - 3 x ^{2} ) ^{2}} : - \frac{28 x}{( 1 - 3 x ^{2} ) ^{2}}

= - \frac{28 x}{( 1 - 3 x ^{2} ) ^{2}}

\frac{l ^{3}}{4} k \frac{1}{4}

3 x + y

\frac{2 x}{( x ^{2} + y ^{2} )}

sec^{3} (\frac{x}{2})

\frac{1}{( 2 x + 2 )}

What is the derivative of (x^2-5)/(1-3x^2) ?
The derivative of (x^2-5)/(1-3x^2) is -(28x)/((1-3x^2)^2)
What is the first derivative of (x^2-5)/(1-3x^2) ?
The first derivative of (x^2-5)/(1-3x^2) is -(28x)/((1-3x^2)^2)