What is (\partial)/(\partial y)(sqrt(x^2+6xy+4y^2)) ?

The answer to (\partial)/(\partial y)(sqrt(x^2+6xy+4y^2)) is (4y+3x)/(sqrt(x^2+6xy+4y^2))

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(\partial)/(\partial y)(sqrt(x^2+6xy+4y^2))

\frac{\partial}{\partial y} (x^{2} + 6 x y + 4 y^{2})

\frac{4 y + 3 x}{x ^{2} + 6 x y + 4 y ^{2}}

Solution steps

\frac{\partial}{\partial y} (x^{2} + 6 x y + 4 y^{2})

Treat

x

as a constant

Apply the chain rule:

\frac{1}{2 x ^{2} + 6 x y + 4 y ^{2}} \frac{\partial}{\partial y} (x^{2} + 6 x y + 4 y^{2})

= \frac{1}{2 x ^{2} + 6 x y + 4 y ^{2}} \frac{\partial}{\partial y} (x^{2} + 6 x y + 4 y^{2})

\frac{\partial}{\partial y} (x^{2} + 6 x y + 4 y^{2}) = 8 y + 6 x

= \frac{1}{2 x ^{2} + 6 x y + 4 y ^{2}} (8 y + 6 x)

Simplify

\frac{1}{2 x ^{2} + 6 x y + 4 y ^{2}} (8 y + 6 x) : \frac{4 y + 3 x}{x ^{2} + 6 x y + 4 y ^{2}}

= \frac{4 y + 3 x}{x ^{2} + 6 x y + 4 y ^{2}}

\int x^{2} + 2 x + 1 d x

\frac{x ^{2} - 1}{2 x - 3}

\int_{0}^{1} \frac{2 y}{e ^{2 y}} d y

\frac{d}{d x} (- (x - 5)^{2})

y^{^{'}} + \frac{3}{t} y = 0

What is (\partial)/(\partial y)(sqrt(x^2+6xy+4y^2)) ?
The answer to (\partial)/(\partial y)(sqrt(x^2+6xy+4y^2)) is (4y+3x)/(sqrt(x^2+6xy+4y^2))