What is the derivative of y=(x^3)/3+1/(4x) ?

The derivative of y=(x^3)/3+1/(4x) is x^2-1/(4x^2)

What is the first derivative of y=(x^3)/3+1/(4x) ?

The first derivative of y=(x^3)/3+1/(4x) is x^2-1/(4x^2)

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derivative of

y = \frac{x ^{3}}{3} + \frac{1}{4 x}

x^{2} - \frac{1}{4 x ^{2}}

Solution steps

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

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{3}}{3}) + \frac{d}{d x} (\frac{1}{4 x})

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

\frac{d}{d x} (\frac{1}{4 x}) = - \frac{1}{4 x ^{2}}

= x^{2} - \frac{1}{4 x ^{2}}

What is the derivative of y=(x^3)/3+1/(4x) ?
The derivative of y=(x^3)/3+1/(4x) is x^2-1/(4x^2)
What is the first derivative of y=(x^3)/3+1/(4x) ?
The first derivative of y=(x^3)/3+1/(4x) is x^2-1/(4x^2)