What is the derivative of (x^2-16/(x-4)) ?

The derivative of (x^2-16/(x-4)) is 1

What is the first derivative of (x^2-16/(x-4)) ?

The first derivative of (x^2-16/(x-4)) is 1

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\frac{d}{d x} (\frac{x ^{2} - 16}{x - 4})

1

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 16}{x - 4})

Simplify

\frac{x ^{2} - 16}{x - 4} : x + 4

= \frac{d}{d x} (x + 4)

Apply the Sum/Difference Rule:

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

= \frac{d x}{d x} + \frac{d}{d x} (4)

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

\frac{d}{d x} (4) = 0

= 1 + 0

Simplify

= 1

\int \frac{x ^{2} + 10 x + 2}{x ^{3} + 15 x ^{2} + 6 x + 9} d x

f (x) = 3 - 2 x

x \to 0 lim (tan (2 x) csc (4 x))

y = x^{4} + 5 e^{x}, (0, 5)

\frac{d y}{d x} = π y + 12 e

What is the derivative of (x^2-16/(x-4)) ?
The derivative of (x^2-16/(x-4)) is 1
What is the first derivative of (x^2-16/(x-4)) ?
The first derivative of (x^2-16/(x-4)) is 1