What is the derivative of (x^3-1/(ln(x))) ?

The derivative of (x^3-1/(ln(x))) is (3x^3ln(x)-x^3+1)/(xln^2(x))

What is the first derivative of (x^3-1/(ln(x))) ?

The first derivative of (x^3-1/(ln(x))) is (3x^3ln(x)-x^3+1)/(xln^2(x))

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (x^{3} - 1) = 3 x^{2}

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

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

Simplify

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

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

\int cos (x) sin (x) 1 + cos^{2} (x) d x

4 y^{^{''}} - 16 y = 0

\int 4 tan (x) d x

f (x) = 2 e^{x}

(ln (3 x + 1))^{^{'}}

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