What is the derivative of f(x)=(x^3)/(6x-3) ?

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

What is the first derivative of f(x)=(x^3)/(6x-3) ?

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

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derivative of

f (x) = \frac{x ^{3}}{6 x - 3}

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

Solution steps

\frac{d}{d x} (\frac{x ^{3}}{6 x - 3})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{3} ) ( 6 x - 3 ) - \frac{d}{d x} ( 6 x - 3 ) x ^{3}}{( 6 x - 3 ) ^{2}}

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

\frac{d}{d x} (6 x - 3) = 6

= \frac{3 x ^{2} ( 6 x - 3 ) - 6 x ^{3}}{( 6 x - 3 ) ^{2}}

Simplify

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

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

x \to 3 + lim (\frac{2 + x}{3 - x})

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

n = 0 \sum \infty n (n + 1)

\int_{4}^{9} \frac{1}{x ^{2} - 1} d x

\int_{0}^{1} 4 e^{7 x} d x

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