What is the derivative of y=((3x^2-x-2))/(3x^2) ?

The derivative of y=((3x^2-x-2))/(3x^2) is (x+4)/(3x^3)

What is the first derivative of y=((3x^2-x-2))/(3x^2) ?

The first derivative of y=((3x^2-x-2))/(3x^2) is (x+4)/(3x^3)

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derivative of

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

\frac{x + 4}{3 x ^{3}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{x + 4}{3 x ^{3}}

\int \frac{cos ^{3} ( x )}{sin ^{4} ( x )} d x

\int_{- 6}^{4} ∣ x ∣ d x

t \to 1 lim (\frac{t - 1}{t ^{2} - 1})

\frac{x ^{3}}{3}

\int cos (\frac{5 x}{3}) d x

What is the derivative of y=((3x^2-x-2))/(3x^2) ?
The derivative of y=((3x^2-x-2))/(3x^2) is (x+4)/(3x^3)
What is the first derivative of y=((3x^2-x-2))/(3x^2) ?
The first derivative of y=((3x^2-x-2))/(3x^2) is (x+4)/(3x^3)