What is the derivative of (1+x^2/x) ?

The derivative of (1+x^2/x) is (x^2-1)/(x^2)

What is the first derivative of (1+x^2/x) ?

The first derivative of (1+x^2/x) is (x^2-1)/(x^2)

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\frac{d}{d x} (\frac{1 + x ^{2}}{x})

\frac{x ^{2} - 1}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{1 + x ^{2}}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 1 + x ^{2} ) x - \frac{d x}{d x} ( 1 + x ^{2} )}{x ^{2}}

\frac{d}{d x} (1 + x^{2}) = 2 x

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

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

Simplify

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

= \frac{x ^{2} - 1}{x ^{2}}

x \to 4 lim (\frac{6}{x})

\int x y^{2} d x

\frac{6 d u}{d t} = u^{2}

\frac{x ^{2} + 2 x}{4 x}

\frac{d y}{d x} = - y + 1

What is the derivative of (1+x^2/x) ?
The derivative of (1+x^2/x) is (x^2-1)/(x^2)
What is the first derivative of (1+x^2/x) ?
The first derivative of (1+x^2/x) is (x^2-1)/(x^2)