What is the derivative of (x^3-6x^2+5/x) ?

The derivative of (x^3-6x^2+5/x) is (2x^3-6x^2-5)/(x^2)

What is the first derivative of (x^3-6x^2+5/x) ?

The first derivative of (x^3-6x^2+5/x) is (2x^3-6x^2-5)/(x^2)

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\frac{d}{d x} (\frac{x ^{3} - 6 x ^{2} + 5}{x})

\frac{2 x ^{3} - 6 x ^{2} - 5}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{3} - 6 x ^{2} + 5}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{3} - 6 x ^{2} + 5 ) x - \frac{d x}{d x} ( x ^{3} - 6 x ^{2} + 5 )}{x ^{2}}

\frac{d}{d x} (x^{3} - 6 x^{2} + 5) = 3 x^{2} - 12 x

\frac{d x}{d x} = 1

= \frac{( 3 x ^{2} - 12 x ) x - 1 \cdot ( x ^{3} - 6 x ^{2} + 5 )}{x ^{2}}

Simplify

\frac{( 3 x ^{2} - 12 x ) x - 1 \cdot ( x ^{3} - 6 x ^{2} + 5 )}{x ^{2}} : \frac{2 x ^{3} - 6 x ^{2} - 5}{x ^{2}}

= \frac{2 x ^{3} - 6 x ^{2} - 5}{x ^{2}}

n = 0 \sum \infty (\frac{1}{6})^{n}

\frac{\partial}{\partial y} (e^{x y})

\frac{\partial}{\partial y} (3 cos (y))

x \to 3 lim (\frac{( x ^{4} - 81 )}{x - 3})

f (x) = (x^{5} - 4 x^{3} - 7)^{8}

What is the derivative of (x^3-6x^2+5/x) ?
The derivative of (x^3-6x^2+5/x) is (2x^3-6x^2-5)/(x^2)
What is the first derivative of (x^3-6x^2+5/x) ?
The first derivative of (x^3-6x^2+5/x) is (2x^3-6x^2-5)/(x^2)