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

(x^{2} + 3) (x^{2} - 3) x^{2}

\int sec^{2} (x) tan^{2} (x) d x

5000 (1 - \frac{1}{40} t)^{2}

\frac{\partial}{\partial x} (- \frac{1}{x - y})

\frac{\partial}{\partial x} (4 x^{3} y^{3} + 1)

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