What is the derivative of (7+8x-4sqrt(x)/x) ?

The derivative of (7+8x-4sqrt(x)/x) is (2sqrt(x)-7)/(x^2)

What is the first derivative of (7+8x-4sqrt(x)/x) ?

The first derivative of (7+8x-4sqrt(x)/x) is (2sqrt(x)-7)/(x^2)

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\frac{d}{d x} (\frac{7 + 8 x - 4 x}{x})

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

Solution steps

\frac{d}{d x} (\frac{7 + 8 x - 4 x}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 7 + 8 x - 4 x ) x - \frac{d x}{d x} ( 7 + 8 x - 4 x )}{x ^{2}}

\frac{d}{d x} (7 + 8 x - 4 x) = - \frac{2}{x} + 8

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

= \frac{( - \frac{2}{x} + 8 ) x - 1 \cdot ( 7 + 8 x - 4 x )}{x ^{2}}

Simplify

\frac{( - \frac{2}{x} + 8 ) x - 1 \cdot ( 7 + 8 x - 4 x )}{x ^{2}} : \frac{2 x - 7}{x ^{2}}

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

\int_{3}^{6} 2 π x (- \frac{3}{2} x + 9) d x

\int \frac{484 - x ^{2}}{x} d x

\int \frac{u}{2 u + 1} d u

x^{3}, x + 6 = y, \frac{- x}{2} = y

\frac{\partial}{\partial y} (ln (x y))

What is the derivative of (7+8x-4sqrt(x)/x) ?
The derivative of (7+8x-4sqrt(x)/x) is (2sqrt(x)-7)/(x^2)
What is the first derivative of (7+8x-4sqrt(x)/x) ?
The first derivative of (7+8x-4sqrt(x)/x) is (2sqrt(x)-7)/(x^2)