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

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

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

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

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derivative of

y = \frac{x ^{2} + 1}{2 x ^{2} - 1}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (2 x^{2} - 1) = 4 x

= \frac{2 x ( 2 x ^{2} - 1 ) - 4 x ( x ^{2} + 1 )}{( 2 x ^{2} - 1 ) ^{2}}

Simplify

\frac{2 x ( 2 x ^{2} - 1 ) - 4 x ( x ^{2} + 1 )}{( 2 x ^{2} - 1 ) ^{2}} : - \frac{6 x}{( 2 x ^{2} - 1 ) ^{2}}

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

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