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}}

f (x) = x^{2} - 5 x + 5, (4, 1)

\int \frac{d}{1} (e^{x} cos (x)) d x

\frac{d}{d x} (\frac{1}{35} x^{5} + \frac{1}{4} x^{4} + x + 1)

f (x) = 8 x 16 + 0.5 x

\int_{2}^{5} \frac{x}{3} 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)