What is the derivative of (x^2-1/(x^2-4)) ?

The derivative of (x^2-1/(x^2-4)) is -(6x)/((x^2-4)^2)

What is the first derivative of (x^2-1/(x^2-4)) ?

The first derivative of (x^2-1/(x^2-4)) is -(6x)/((x^2-4)^2)

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= - \frac{6 x}{( x ^{2} - 4 ) ^{2}}

\int \frac{cos ( y )}{sin ^{2} ( y )} d y

\frac{\partial}{\partial x} (5 e^{y})

n = 1 \sum \infty \frac{1}{( n ^{2} + 8 n )}

\int \frac{1}{x ( 1 + ln ( x ) ) ( 2 + ln ( x ) )} d x

\int \frac{8 + 5 x}{1 + x ^{2}} d x

What is the derivative of (x^2-1/(x^2-4)) ?
The derivative of (x^2-1/(x^2-4)) is -(6x)/((x^2-4)^2)
What is the first derivative of (x^2-1/(x^2-4)) ?
The first derivative of (x^2-1/(x^2-4)) is -(6x)/((x^2-4)^2)