What is (\partial)/(\partial x)(sqrt(x^2+6xy+y^2)) ?

The answer to (\partial)/(\partial x)(sqrt(x^2+6xy+y^2)) is (x+3y)/(sqrt(x^2+6xy+y^2))

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(\partial)/(\partial x)(sqrt(x^2+6xy+y^2))

\frac{\partial}{\partial x} (x^{2} + 6 x y + y^{2})

\frac{x + 3 y}{x ^{2} + 6 x y + y ^{2}}

Solution steps

\frac{\partial}{\partial x} (x^{2} + 6 x y + y^{2})

Treat

y

as a constant

Apply the chain rule:

\frac{1}{2 x ^{2} + 6 x y + y ^{2}} \frac{\partial}{\partial x} (x^{2} + 6 x y + y^{2})

= \frac{1}{2 x ^{2} + 6 x y + y ^{2}} \frac{\partial}{\partial x} (x^{2} + 6 x y + y^{2})

\frac{\partial}{\partial x} (x^{2} + 6 x y + y^{2}) = 2 x + 6 y

= \frac{1}{2 x ^{2} + 6 x y + y ^{2}} (2 x + 6 y)

Simplify

\frac{1}{2 x ^{2} + 6 x y + y ^{2}} (2 x + 6 y) : \frac{x + 3 y}{x ^{2} + 6 x y + y ^{2}}

= \frac{x + 3 y}{x ^{2} + 6 x y + y ^{2}}

(x^{2} + 1) \frac{d y}{d x} + 3 x (y - 1) = 0, y (0) = 9

\frac{\partial}{\partial y} (2 x^{2} + 4 y^{2} + \frac{1}{2})

y^{^{'}} = 1 + y

\int_{0}^{1} \frac{1}{x - 1} d x

y^{^{'}} = \frac{( x y + 3 x - y - 3 )}{( x y - 2 x + 4 y - 8 )}

What is (\partial)/(\partial x)(sqrt(x^2+6xy+y^2)) ?
The answer to (\partial)/(\partial x)(sqrt(x^2+6xy+y^2)) is (x+3y)/(sqrt(x^2+6xy+y^2))