What is the derivative of ({g}(x)/(sqrt(x))) ?

The derivative of ({g}(x)/(sqrt(x))) is (2x derivative of {g}(x)-{g}(x))/(2xsqrt(x))

What is the first derivative of ({g}(x)/(sqrt(x))) ?

The first derivative of ({g}(x)/(sqrt(x))) is (2x derivative of {g}(x)-{g}(x))/(2xsqrt(x))

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

\frac{2 x \frac{d}{d x} ( g ( x ) ) - g ( x )}{2 x x}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

Simplify

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

= \frac{2 x \frac{d}{d x} ( g ( x ) ) - g ( x )}{2 x x}

\int - 4 y^{2} d y

\frac{\partial}{\partial x} ((x + y)^{3})

\frac{\partial}{\partial x} (x^{2} yz - x y z^{3})

y = 7 x, y = x^{2} - 8

f (x) = \frac{x}{x + 9}

What is the derivative of ({g}(x)/(sqrt(x))) ?
The derivative of ({g}(x)/(sqrt(x))) is (2x derivative of {g}(x)-{g}(x))/(2xsqrt(x))
What is the first derivative of ({g}(x)/(sqrt(x))) ?
The first derivative of ({g}(x)/(sqrt(x))) is (2x derivative of {g}(x)-{g}(x))/(2xsqrt(x))