What is the derivative of (x^4+3/(6x)) ?

The derivative of (x^4+3/(6x)) is (x^4-1)/(2x^2)

What is the first derivative of (x^4+3/(6x)) ?

The first derivative of (x^4+3/(6x)) is (x^4-1)/(2x^2)

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\frac{d}{d x} (\frac{x ^{4} + 3}{6 x})

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

Solution steps

\frac{d}{d x} (\frac{x ^{4} + 3}{6 x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{6} \frac{d}{d x} (\frac{x ^{4} + 3}{x})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{6} \cdot \frac{\frac{d}{d x} ( x ^{4} + 3 ) x - \frac{d x}{d x} ( x ^{4} + 3 )}{x ^{2}}

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

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

= \frac{1}{6} \cdot \frac{4 x ^{3} x - 1 \cdot ( x ^{4} + 3 )}{x ^{2}}

Simplify

\frac{1}{6} \cdot \frac{4 x ^{3} x - 1 \cdot ( x ^{4} + 3 )}{x ^{2}} : \frac{x ^{4} - 1}{2 x ^{2}}

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

\frac{d}{d y} (3 y)

\frac{d}{d x} (1 + 2 e^{3 x})

f (y) = 2 x, 0, 4

e^{t} cos (t) + e^{t} sin (t)

\frac{d}{d x} (arctan (sin (2 x)))

What is the derivative of (x^4+3/(6x)) ?
The derivative of (x^4+3/(6x)) is (x^4-1)/(2x^2)
What is the first derivative of (x^4+3/(6x)) ?
The first derivative of (x^4+3/(6x)) is (x^4-1)/(2x^2)