What is the derivative of (x^3/(48)+4/x) ?

The derivative of (x^3/(48)+4/x) is (x^2)/(16)-4/(x^2)

What is the first derivative of (x^3/(48)+4/x) ?

The first derivative of (x^3/(48)+4/x) is (x^2)/(16)-4/(x^2)

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\frac{d}{d x} (\frac{x ^{3}}{48} + \frac{4}{x})

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

Solution steps

\frac{d}{d x} (\frac{x ^{3}}{48} + \frac{4}{x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{3}}{48}) + \frac{d}{d x} (\frac{4}{x})

\frac{d}{d x} (\frac{x ^{3}}{48}) = \frac{x ^{2}}{16}

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

= \frac{x ^{2}}{16} - \frac{4}{x ^{2}}

\int \frac{1}{sin ( x ) tan ( x )} d x

\int \frac{17}{x} d x

\frac{1}{a s + 1}

x \to 1 lim (\frac{I n x}{x - 1})

\int ln (k x) d x

What is the derivative of (x^3/(48)+4/x) ?
The derivative of (x^3/(48)+4/x) is (x^2)/(16)-4/(x^2)
What is the first derivative of (x^3/(48)+4/x) ?
The first derivative of (x^3/(48)+4/x) is (x^2)/(16)-4/(x^2)