What is (\partial)/(\partial x)(((x+1))/(x^2)) ?

The answer to (\partial)/(\partial x)(((x+1))/(x^2)) is (-x-2)/(x^3)

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(\partial)/(\partial x)(((x+1))/(x^2))

\frac{\partial}{\partial x} (\frac{( x + 1 )}{x ^{2}})

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

Solution steps

\frac{\partial}{\partial x} (\frac{( x + 1 )}{x ^{2}})

Apply the Quotient Rule:

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

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

\frac{\partial}{\partial x} (x + 1) = 1

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

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

Simplify

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

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

\frac{\partial}{\partial x} (\frac{2}{x ^{2} + y ^{2} + z ^{2}})

\int 20 x (5 x^{2} - 3)^{3} d x

\int (sin^{2} (x))^{2} d x

\frac{d ^{2} y}{d x ^{2}} + 4 y = 0

(f (x) (2 x))^{'}

What is (\partial)/(\partial x)(((x+1))/(x^2)) ?
The answer to (\partial)/(\partial x)(((x+1))/(x^2)) is (-x-2)/(x^3)