What is the derivative of (x^2-8x+4/(2x)) ?

The derivative of (x^2-8x+4/(2x)) is (x^2-4)/(2x^2)

What is the first derivative of (x^2-8x+4/(2x)) ?

The first derivative of (x^2-8x+4/(2x)) is (x^2-4)/(2x^2)

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Popular >

\frac{d}{d x} (\frac{x ^{2} - 8 x + 4}{2 x})

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

Solution steps

\frac{d}{d x} (\frac{x ^{2} - 8 x + 4}{2 x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{2} \frac{d}{d x} (\frac{x ^{2} - 8 x + 4}{x})

Apply the Quotient Rule:

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

= \frac{1}{2} \cdot \frac{\frac{d}{d x} ( x ^{2} - 8 x + 4 ) x - \frac{d x}{d x} ( x ^{2} - 8 x + 4 )}{x ^{2}}

\frac{d}{d x} (x^{2} - 8 x + 4) = 2 x - 8

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

= \frac{1}{2} \cdot \frac{( 2 x - 8 ) x - 1 \cdot ( x ^{2} - 8 x + 4 )}{x ^{2}}

Simplify

\frac{1}{2} \cdot \frac{( 2 x - 8 ) x - 1 \cdot ( x ^{2} - 8 x + 4 )}{x ^{2}} : \frac{x ^{2} - 4}{2 x ^{2}}

= \frac{x ^{2} - 4}{2 x ^{2}}

x \to 2 lim (3 - x^{2})

(1 + 7 x^{2}) (x - x^{2})

\frac{d y}{d x} + (\frac{7}{x}) y = x + 7

arccos (e^{3 x})

\int_{0}^{\frac{π}{4}} tan^{2} (x) d x

What is the derivative of (x^2-8x+4/(2x)) ?
The derivative of (x^2-8x+4/(2x)) is (x^2-4)/(2x^2)
What is the first derivative of (x^2-8x+4/(2x)) ?
The first derivative of (x^2-8x+4/(2x)) is (x^2-4)/(2x^2)