What is the derivative of (2x+7/(2x)) ?

The derivative of (2x+7/(2x)) is -7/(2x^2)

What is the first derivative of (2x+7/(2x)) ?

The first derivative of (2x+7/(2x)) is -7/(2x^2)

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\frac{d}{d x} (\frac{2 x + 7}{2 x})

- \frac{7}{2 x ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 x + 7}{2 x})

Take the constant out:

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

= \frac{1}{2} \frac{d}{d x} (\frac{2 x + 7}{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} ( 2 x + 7 ) x - \frac{d x}{d x} ( 2 x + 7 )}{x ^{2}}

\frac{d}{d x} (2 x + 7) = 2

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

= \frac{1}{2} \cdot \frac{2 x - 1 \cdot ( 2 x + 7 )}{x ^{2}}

Simplify

\frac{1}{2} \cdot \frac{2 x - 1 \cdot ( 2 x + 7 )}{x ^{2}} : - \frac{7}{2 x ^{2}}

= - \frac{7}{2 x ^{2}}

- (- x^{3} + y^{3}) d x + x y^{2} d y = 0

\int \frac{1}{3 x ^{2} + 4 x - 7} d x

\frac{d}{d x} (x (x^{5} + 9)^{3})

y = x^{4} - 5 x^{3} + 3 x^{2} - 4 x, (- 1, 13)

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

What is the derivative of (2x+7/(2x)) ?
The derivative of (2x+7/(2x)) is -7/(2x^2)
What is the first derivative of (2x+7/(2x)) ?
The first derivative of (2x+7/(2x)) is -7/(2x^2)