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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d y}{d t} = \frac{y}{t + 1} + 4 t^{2} + 4 t, y (1) = - 5

x \to 0 lim (\frac{1}{x} - 3)

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

\int (x + 1) 4 x + 5 d x

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

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