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

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

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

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

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derivative of

y = \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}}

ln (x + 6)

y^{^{'}} = 9 y^{2} + x y^{2}

\frac{d}{d x} (\frac{2 x}{x - 1})

y = e^{4} x^{2} + 18

f (x) = (\frac{8 x + 7}{9 x ^{2} + 6})

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