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

The derivative of (sqrt(x)+x)/(x^2) is (-2x-3sqrt(x))/(2x^3)

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

The first derivative of (sqrt(x)+x)/(x^2) is (-2x-3sqrt(x))/(2x^3)

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derivative of

\frac{x + x}{x ^{2}}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (ln (4 + x))

\int x e^{x} - e^{x} d x

\frac{d y}{d x} = 2 x y + \frac{6}{π}

\int \frac{1}{x ^{4} 9 - x ^{2}} d x

n = 1 \sum \infty 1 - \frac{3}{7 ^{n}}

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