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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int_{0}^{1} x e^{- 8 x^{2}} d x

\frac{d y}{d x} + 9 y = 3

\int \frac{14}{1 - cos ( 7 x )} d x

y^{^{''}} + 2 y = cos (t) - 2 \cdot sin (2 t)

\int 3 e^{- 3 x} d x

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