What is the derivative of f(x)=(x^2-2x-3)/x ?

The derivative of f(x)=(x^2-2x-3)/x is (x^2+3)/(x^2)

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

The first derivative of f(x)=(x^2-2x-3)/x is (x^2+3)/(x^2)

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derivative of

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

\frac{x ^{2} + 3}{x ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

= \frac{( 2 x - 2 ) x - 1 \cdot ( x ^{2} - 2 x - 3 )}{x ^{2}}

Simplify

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

= \frac{x ^{2} + 3}{x ^{2}}

\frac{\partial}{\partial x} (3 x^{2} - 12 y)

\frac{d}{d x} (\frac{18}{x ^{2} + 2})

\frac{4}{x - 4} + \frac{27}{2 x + 2}

x^{^{'}} - x = 0

\frac{\partial}{\partial x} (ln (x^{2} + y x))

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