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

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

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

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

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derivative of

y = \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}}

\int (2 x - \frac{3}{x})^{2} d x

(- 7, - 9), (- 2, - 9)

\int (x^{1.4} + 7 x^{2.5}) d x

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

f (x) = \frac{x}{x ^{2} + 4}

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