What is the derivative of (9sqrt(x)/x) ?

The derivative of (9sqrt(x)/x) is -9/(2x^{3/2)}

What is the first derivative of (9sqrt(x)/x) ?

The first derivative of (9sqrt(x)/x) is -9/(2x^{3/2)}

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\frac{d}{d x} (\frac{9 x}{x})

- \frac{9}{2 x ^{\frac{3}{2}}}

Solution steps

\frac{d}{d x} (\frac{9 x}{x})

Take the constant out:

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

= 9 \frac{d}{d x} (\frac{x}{x})

Apply the Quotient Rule:

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

= 9 \cdot \frac{\frac{d}{d x} ( x ) x - \frac{d x}{d x} x}{x ^{2}}

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

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

= 9 \cdot \frac{\frac{1}{2 x} x - 1 \cdot x}{x ^{2}}

Simplify

9 \cdot \frac{\frac{1}{2 x} x - 1 \cdot x}{x ^{2}} : - \frac{9}{2 x ^{\frac{3}{2}}}

= - \frac{9}{2 x ^{\frac{3}{2}}}

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

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

\frac{1 - 4 t}{3 + t}

\int \frac{( 2 z - 1 )}{2 z ^{2} - 6 z + 5} d z

(9 x + 1) y^{2} \frac{d y}{d x} + 4 x^{2} + 3 y^{3} = 0

What is the derivative of (9sqrt(x)/x) ?
The derivative of (9sqrt(x)/x) is -9/(2x^{3/2)}
What is the first derivative of (9sqrt(x)/x) ?
The first derivative of (9sqrt(x)/x) is -9/(2x^{3/2)}