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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (x + 9) = 1

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

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

Simplify

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

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

\frac{d}{d x} (\frac{3 x ^{4} + 3}{sin ( x )})

\frac{d}{d s} (s^{2} + t^{2})

3 x^{4} - 7 x^{2}, at x = - 1

\int \frac{- 2 x + 4}{( x ^{2} + 1 ) ( x - 1 ) ^{2}} d x

\frac{d y}{d x} = 4 y^{2} sec^{2} (2 x), y (\frac{π}{8}) = 1

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