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

The derivative of (2x^2-1/(x^2-3)) is -(10x)/((x^2-3)^2)

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

The first derivative of (2x^2-1/(x^2-3)) is -(10x)/((x^2-3)^2)

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

- \frac{10 x}{( x ^{2} - 3 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

= \frac{4 x ( x ^{2} - 3 ) - 2 x ( 2 x ^{2} - 1 )}{( x ^{2} - 3 ) ^{2}}

Simplify

\frac{4 x ( x ^{2} - 3 ) - 2 x ( 2 x ^{2} - 1 )}{( x ^{2} - 3 ) ^{2}} : - \frac{10 x}{( x ^{2} - 3 ) ^{2}}

= - \frac{10 x}{( x ^{2} - 3 ) ^{2}}

\int 20 x^{3} + 12 x^{2} + 4 d x

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

\frac{d y}{d x} + y^{3} x + 5 y = 0

6 x (\frac{d y}{d x}) = 2 x e^{x} - 6 y + 6 x^{2}

\int_{0}^{1} 4 x y d y

What is the derivative of (2x^2-1/(x^2-3)) ?
The derivative of (2x^2-1/(x^2-3)) is -(10x)/((x^2-3)^2)
What is the first derivative of (2x^2-1/(x^2-3)) ?
The first derivative of (2x^2-1/(x^2-3)) is -(10x)/((x^2-3)^2)