What is the derivative of y=(9+xf(x))/(sqrt(x)) ?

The derivative of y=(9+xf(x))/(sqrt(x)) is (2x^2\frac{df)/(dx)+xf(x)-9}{2xsqrt(x)}

What is the first derivative of y=(9+xf(x))/(sqrt(x)) ?

The first derivative of y=(9+xf(x))/(sqrt(x)) is (2x^2\frac{df)/(dx)+xf(x)-9}{2xsqrt(x)}

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derivative of

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

\frac{2 x ^{2} \frac{df}{d x} + x f ( x ) - 9}{2 x x}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

= \frac{( x \frac{df}{d x} + f ( x ) ) x - \frac{1}{2 x} ( 9 + x f ( x ) )}{( x ) ^{2}}

Simplify

\frac{( x \frac{df}{d x} + f ( x ) ) x - \frac{1}{2 x} ( 9 + x f ( x ) )}{( x ) ^{2}} : \frac{2 x ^{2} \frac{df}{d x} + x f ( x ) - 9}{2 x x}

= \frac{2 x ^{2} \frac{df}{d x} + x f ( x ) - 9}{2 x x}

\frac{d}{d x} (3 x^{2} sin (x))

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

\frac{1}{x} sin (x)

f (x) = x^{4} (3 - x)^{3}, at x = 1

\frac{12 x}{x ^{2} - 4}

What is the derivative of y=(9+xf(x))/(sqrt(x)) ?
The derivative of y=(9+xf(x))/(sqrt(x)) is (2x^2\frac{df)/(dx)+xf(x)-9}{2xsqrt(x)}
What is the first derivative of y=(9+xf(x))/(sqrt(x)) ?
The first derivative of y=(9+xf(x))/(sqrt(x)) is (2x^2\frac{df)/(dx)+xf(x)-9}{2xsqrt(x)}