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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

c cos (t) + t^{2} sin (t)

\int (2 + 2 cos (θ))^{2} d θ

\int \frac{1}{4 x ^{2} + 4 x + 10} d x

\frac{d y}{d x} = (3 x^{2} y)

4 x, 8 x^{2}

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