What is the derivative of f(x)=(x^2+9)/(sqrt(x)) ?

The derivative of f(x)=(x^2+9)/(sqrt(x)) is (3x^2-9)/(2xsqrt(x))

What is the first derivative of f(x)=(x^2+9)/(sqrt(x)) ?

The first derivative of f(x)=(x^2+9)/(sqrt(x)) is (3x^2-9)/(2xsqrt(x))

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{2 x x - \frac{1}{2 x} ( x ^{2} + 9 )}{( x ) ^{2}} : \frac{3 x ^{2} - 9}{2 x x}

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

What is the derivative of f(x)=(x^2+9)/(sqrt(x)) ?
The derivative of f(x)=(x^2+9)/(sqrt(x)) is (3x^2-9)/(2xsqrt(x))
What is the first derivative of f(x)=(x^2+9)/(sqrt(x)) ?
The first derivative of f(x)=(x^2+9)/(sqrt(x)) is (3x^2-9)/(2xsqrt(x))