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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 1 - \frac{16}{x ^{2}}

f (x) = 24 t^{\frac{3}{2}} + 4 t^{\frac{1}{2}}

\int 2 - y^{2} d y

\int_{- 2}^{4} (1 + 2 x - 3 x^{2} + x^{3}) d x

\int \frac{1}{y ^{2} + y} d y

\frac{x}{x ^{2} + 49}

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