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

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

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

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = (x^{2} + 1)^{- 1}

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

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

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

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

Simplify

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

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

e^{4 x}

\int_{0}^{5} \frac{1}{x ^{\frac{15}{16}}} d x

y = 6 x^{2} ln (x), y = 24 ln (x)

\frac{\partial}{\partial z} (x z - e^{x} sin (y))

f (x) = tan (3 x)

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