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

The derivative of (3x^4-x/x) is 9x^2

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

The first derivative of (3x^4-x/x) is 9x^2

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\frac{d}{d x} (\frac{3 x ^{4} - x}{x})

9 x^{2}

Solution steps

\frac{d}{d x} (\frac{3 x ^{4} - x}{x})

Simplify

\frac{3 x ^{4} - x}{x} : 3 x^{3} - 1

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (3 x^{3}) = 9 x^{2}

\frac{d}{d x} (1) = 0

= 9 x^{2} - 0

Simplify

= 9 x^{2}

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What is the derivative of (3x^4-x/x) ?
The derivative of (3x^4-x/x) is 9x^2
What is the first derivative of (3x^4-x/x) ?
The first derivative of (3x^4-x/x) is 9x^2