What is the derivative of (x^2-64/(x+8)) ?

The derivative of (x^2-64/(x+8)) is 1

What is the first derivative of (x^2-64/(x+8)) ?

The first derivative of (x^2-64/(x+8)) is 1

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\frac{d}{d x} (\frac{x ^{2} - 64}{x + 8})

1

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 64}{x + 8})

Simplify

\frac{x ^{2} - 64}{x + 8} : x - 8

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d x}{d x} = 1

\frac{d}{d x} (8) = 0

= 1 - 0

Simplify

= 1

g (t) = (9 t + 5)^{2} (5 t - 6)^{- 3}

(3 x + 2) + (3 y - 3) y^{^{'}} = 0

\int y (y + 1) e^{- 2 y} d y

\int e + e^{3 x} d x

\frac{d}{d x} (x^{- 2} e^{2 x})

What is the derivative of (x^2-64/(x+8)) ?
The derivative of (x^2-64/(x+8)) is 1
What is the first derivative of (x^2-64/(x+8)) ?
The first derivative of (x^2-64/(x+8)) is 1