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

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

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

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

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\frac{d}{d x} (x^{2} + \frac{16}{x})

2 x - \frac{16}{x ^{2}}

Solution steps

\frac{d}{d x} (x^{2} + \frac{16}{x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{2}) + \frac{d}{d x} (\frac{16}{x})

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

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

= 2 x - \frac{16}{x ^{2}}

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 1

\frac{d}{d x} ((x + 1)^{s i n (x)})

2 \frac{d y}{d x} = y - e^{4 x}

f (x) = 6 x (6.6)

\frac{\partial}{\partial y} (x sin (x + 2 y))

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