What is the derivative of x+(64/x) ?

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

What is the first derivative of x+(64/x) ?

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

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\frac{d}{d x} (x + \frac{64}{x})

1 - \frac{64}{x ^{2}}

Solution steps

\frac{d}{d x} (x + \frac{64}{x})

Apply the Sum/Difference Rule:

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

= \frac{d x}{d x} + \frac{d}{d x} (\frac{64}{x})

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

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

= 1 - \frac{64}{x ^{2}}

\int \frac{3}{x ^{2} ( x ^{2} + 25 )} d x

\int \frac{1}{36 + x ^{2}} d x

\frac{27}{x ^{2} + 9}

\frac{\partial}{\partial x} (- 2 e^{4 y - x^{2} - y^{2}} x)

x \to \infty lim (x - 6)

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