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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{( 2 + ln ( 7 x ) )}{3 x} d x

6 x^{2} - 48 x + y^{2} + 5 = 0

x \to (- k + 1) lim (\frac{1}{x - 2})

y^{^{'}} + 8 y = 0

\int (x + 1) x^{2} + 2 x + 3 d x

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