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

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

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

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

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derivative of

g (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

y^{^{''}} + 6 y^{^{'}} + 5 y = - 3 t e^{(4 t)}

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

\int 5 cos (x) - 4 sin (x) d x

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