What is the derivative of ((x-3)(x^2+3x+9))/(x^3) ?

The derivative of ((x-3)(x^2+3x+9))/(x^3) is (81)/(x^4)

What is the first derivative of ((x-3)(x^2+3x+9))/(x^3) ?

The first derivative of ((x-3)(x^2+3x+9))/(x^3) is (81)/(x^4)

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derivative of

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

\frac{81}{x ^{4}}

Solution steps

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

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

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

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

Simplify

\frac{3 x ^{2} x ^{3} - 3 x ^{2} ( x - 3 ) ( x ^{2} + 3 x + 9 )}{( x ^{3} ) ^{2}} : \frac{81}{x ^{4}}

= \frac{81}{x ^{4}}

x^{2} y \frac{d y}{d x} - 9 = 0

\int - 13 cos (x) d x

\frac{d}{d r} (\frac{4}{3} π r^{3})

y^{^{'}} - 2 y = e^{2 t}, y (0) = 2

y^{^{'}} + y = 0

What is the derivative of ((x-3)(x^2+3x+9))/(x^3) ?
The derivative of ((x-3)(x^2+3x+9))/(x^3) is (81)/(x^4)
What is the first derivative of ((x-3)(x^2+3x+9))/(x^3) ?
The first derivative of ((x-3)(x^2+3x+9))/(x^3) is (81)/(x^4)