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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\frac{d}{d x} (\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}}

\frac{d y}{d x} = 4 y^{2} - 3 y + 1

x = y + 12, x = y^{2} + 2 y

sin (tan (7 x))

\int_{2}^{4} (10 + 4 x - 3 x^{3}) d x

f (x) = 30 e^{x^{2}} x

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)