What is the derivative of (x^4+2/(ln(x))) ?

The derivative of (x^4+2/(ln(x))) is (4x^4ln(x)-x^4-2)/(xln^2(x))

What is the first derivative of (x^4+2/(ln(x))) ?

The first derivative of (x^4+2/(ln(x))) is (4x^4ln(x)-x^4-2)/(xln^2(x))

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\frac{d}{d x} (\frac{x ^{4} + 2}{ln ( x )})

\frac{4 x ^{4} ln ( x ) - x ^{4} - 2}{x ln ^{2} ( x )}

Solution steps

\frac{d}{d x} (\frac{x ^{4} + 2}{ln ( x )})

Apply the Quotient Rule:

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

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

\frac{d}{d x} (x^{4} + 2) = 4 x^{3}

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

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

Simplify

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

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

\int \frac{x}{e ^{4 x}} d x

n = 1 \sum \infty \frac{2}{9 ^{n}}

3 x - 2 x^{2}

x \to 0 lim ((e^{- \frac{1}{x}})^{x})

\frac{d y}{d t} + t y = t y^{2}

What is the derivative of (x^4+2/(ln(x))) ?
The derivative of (x^4+2/(ln(x))) is (4x^4ln(x)-x^4-2)/(xln^2(x))
What is the first derivative of (x^4+2/(ln(x))) ?
The first derivative of (x^4+2/(ln(x))) is (4x^4ln(x)-x^4-2)/(xln^2(x))