What is the derivative of y=(3+4x-6sqrt(x))/x ?

The derivative of y=(3+4x-6sqrt(x))/x is (3sqrt(x)-3)/(x^2)

What is the first derivative of y=(3+4x-6sqrt(x))/x ?

The first derivative of y=(3+4x-6sqrt(x))/x is (3sqrt(x)-3)/(x^2)

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derivative of

y = \frac{3 + 4 x - 6 x}{x}

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

Solution steps

\frac{d}{d x} (\frac{3 + 4 x - 6 x}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 3 + 4 x - 6 x ) x - \frac{d x}{d x} ( 3 + 4 x - 6 x )}{x ^{2}}

\frac{d}{d x} (3 + 4 x - 6 x) = - \frac{3}{x} + 4

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

= \frac{( - \frac{3}{x} + 4 ) x - 1 \cdot ( 3 + 4 x - 6 x )}{x ^{2}}

Simplify

\frac{( - \frac{3}{x} + 4 ) x - 1 \cdot ( 3 + 4 x - 6 x )}{x ^{2}} : \frac{3 x - 3}{x ^{2}}

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

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

\frac{\partial}{\partial z} (x sin (y - z))

\int \frac{x ^{3}}{u} d x

x^{3} - 3 x^{2} + 3

\int_{0}^{17} x d x

What is the derivative of y=(3+4x-6sqrt(x))/x ?
The derivative of y=(3+4x-6sqrt(x))/x is (3sqrt(x)-3)/(x^2)
What is the first derivative of y=(3+4x-6sqrt(x))/x ?
The first derivative of y=(3+4x-6sqrt(x))/x is (3sqrt(x)-3)/(x^2)