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

The derivative of f(x)=(6x^3)/(4-x) is (6(12x^2-2x^3))/((4-x)^2)

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

The first derivative of f(x)=(6x^3)/(4-x) is (6(12x^2-2x^3))/((4-x)^2)

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derivative of

f (x) = \frac{6 x ^{3}}{4 - x}

\frac{6 ( 12 x ^{2} - 2 x ^{3} )}{( 4 - x ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (4 - x) = - 1

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

Simplify

6 \cdot \frac{3 x ^{2} ( 4 - x ) - ( - 1 ) x ^{3}}{( 4 - x ) ^{2}} : \frac{6 ( 12 x ^{2} - 2 x ^{3} )}{( 4 - x ) ^{2}}

= \frac{6 ( 12 x ^{2} - 2 x ^{3} )}{( 4 - x ) ^{2}}

x \to 0 lim (\frac{3 ^{x^{2}} - 1}{x})

\frac{1}{x + 6}

\int (- 2 x + 3) d x

3 x^{2} + 8

\int \frac{1}{v ^{2} - v} d v

What is the derivative of f(x)=(6x^3)/(4-x) ?
The derivative of f(x)=(6x^3)/(4-x) is (6(12x^2-2x^3))/((4-x)^2)
What is the first derivative of f(x)=(6x^3)/(4-x) ?
The first derivative of f(x)=(6x^3)/(4-x) is (6(12x^2-2x^3))/((4-x)^2)