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

The derivative of y=((x-2)(x^2+2x+4))/(x^3) is (24)/(x^4)

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

The first derivative of y=((x-2)(x^2+2x+4))/(x^3) is (24)/(x^4)

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derivative of

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

\frac{24}{x ^{4}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{3 x ^{2} x ^{3} - 3 x ^{2} ( x - 2 ) ( x ^{2} + 2 x + 4 )}{( x ^{3} ) ^{2}} : \frac{24}{x ^{4}}

= \frac{24}{x ^{4}}

x^{2} + y^{2}

\int_{0}^{1} 2 π (y) (y - y^{3}) d y

\int y^{6} (cos (3 y)) d y

\int (\frac{1}{x})^{2} d x

\frac{1}{( s ^{4} ) ( s - 5 )}

What is the derivative of y=((x-2)(x^2+2x+4))/(x^3) ?
The derivative of y=((x-2)(x^2+2x+4))/(x^3) is (24)/(x^4)
What is the first derivative of y=((x-2)(x^2+2x+4))/(x^3) ?
The first derivative of y=((x-2)(x^2+2x+4))/(x^3) is (24)/(x^4)