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

The derivative of (x^2+1/(x^2-4)) is -(10x)/((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 -(10x)/((x^2-4)^2)

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

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

= - \frac{10 x}{( x ^{2} - 4 ) ^{2}}

(1 + ln (x) + \frac{y}{x}) d x + (- 1 + ln (x)) d y = 0

y = e^{3 x} + 10 e^{2 x}

x \to 0 lim (\frac{2 e ^{x} - 2}{3 x})

t e^{\frac{- t}{2}}

\int \frac{1}{x ( x ^{2} - 81 ) ^{\frac{3}{2}}} d x

What is the derivative of (x^2+1/(x^2-4)) ?
The derivative of (x^2+1/(x^2-4)) is -(10x)/((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 -(10x)/((x^2-4)^2)