What is the derivative of y=(((x^2+8))/(x^2-8)) ?

The derivative of y=(((x^2+8))/(x^2-8)) is -(32x)/((x^2-8)^2)

What is the first derivative of y=(((x^2+8))/(x^2-8)) ?

The first derivative of y=(((x^2+8))/(x^2-8)) is -(32x)/((x^2-8)^2)

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derivative of

y = (\frac{( x ^{2} + 8 )}{x ^{2} - 8})

- \frac{32 x}{( x ^{2} - 8 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} + 8}{x ^{2} - 8})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{2} + 8 ) ( x ^{2} - 8 ) - \frac{d}{d x} ( x ^{2} - 8 ) ( x ^{2} + 8 )}{( x ^{2} - 8 ) ^{2}}

\frac{d}{d x} (x^{2} + 8) = 2 x

\frac{d}{d x} (x^{2} - 8) = 2 x

= \frac{2 x ( x ^{2} - 8 ) - 2 x ( x ^{2} + 8 )}{( x ^{2} - 8 ) ^{2}}

Simplify

\frac{2 x ( x ^{2} - 8 ) - 2 x ( x ^{2} + 8 )}{( x ^{2} - 8 ) ^{2}} : - \frac{32 x}{( x ^{2} - 8 ) ^{2}}

= - \frac{32 x}{( x ^{2} - 8 ) ^{2}}

d er i v a t i v e \frac{1}{( 1 - t ^{2} ) ^{\frac{1}{2}}}

t an g e n t f (x) = (\frac{x + 1}{x + 3})^{2}, at x = - 2

\int_{0}^{\infty} \frac{( y e ^{- \frac{y}{a}} )}{a} d y

a re a y = x^{2} - 4 x, y = 2 x + 7

\int 3 ln (x^{2} - x + 4) d x

What is the derivative of y=(((x^2+8))/(x^2-8)) ?
The derivative of y=(((x^2+8))/(x^2-8)) is -(32x)/((x^2-8)^2)
What is the first derivative of y=(((x^2+8))/(x^2-8)) ?
The first derivative of y=(((x^2+8))/(x^2-8)) is -(32x)/((x^2-8)^2)