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

The derivative of h(x)=(3x^3-3)/(ln(x)) is (9x^3ln(x)-3x^3+3)/(xln^2(x))

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

The first derivative of h(x)=(3x^3-3)/(ln(x)) is (9x^3ln(x)-3x^3+3)/(xln^2(x))

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derivative of

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

\frac{9 x ^{3} ln ( x ) - 3 x ^{3} + 3}{x ln ^{2} ( x )}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

= \frac{9 x ^{2} ln ( x ) - \frac{1}{x} ( 3 x ^{3} - 3 )}{( ln ( x ) ) ^{2}}

Simplify

\frac{9 x ^{2} ln ( x ) - \frac{1}{x} ( 3 x ^{3} - 3 )}{ln ^{2} ( x )} : \frac{9 x ^{3} ln ( x ) - 3 x ^{3} + 3}{x ln ^{2} ( x )}

= \frac{9 x ^{3} ln ( x ) - 3 x ^{3} + 3}{x ln ^{2} ( x )}

f (x) = 1 - 2 x + x^{2}

f (T) = T

y = \frac{3}{x}, y = 9 x, y = \frac{1}{9 x}

4 sin (3 t)

What is the derivative of h(x)=(3x^3-3)/(ln(x)) ?
The derivative of h(x)=(3x^3-3)/(ln(x)) is (9x^3ln(x)-3x^3+3)/(xln^2(x))
What is the first derivative of h(x)=(3x^3-3)/(ln(x)) ?
The first derivative of h(x)=(3x^3-3)/(ln(x)) is (9x^3ln(x)-3x^3+3)/(xln^2(x))