What is (\partial)/(\partial y)(sqrt(5x^2+8xy+y^2)) ?

The answer to (\partial)/(\partial y)(sqrt(5x^2+8xy+y^2)) is (y+4x)/(sqrt(5x^2+8xy+y^2))

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(\partial)/(\partial y)(sqrt(5x^2+8xy+y^2))

\frac{\partial}{\partial y} (5 x^{2} + 8 x y + y^{2})

\frac{y + 4 x}{5 x ^{2} + 8 x y + y ^{2}}

Solution steps

\frac{\partial}{\partial y} (5 x^{2} + 8 x y + y^{2})

Treat

x

as a constant

Apply the chain rule:

\frac{1}{2 5 x ^{2} + 8 x y + y ^{2}} \frac{\partial}{\partial y} (5 x^{2} + 8 x y + y^{2})

= \frac{1}{2 5 x ^{2} + 8 x y + y ^{2}} \frac{\partial}{\partial y} (5 x^{2} + 8 x y + y^{2})

\frac{\partial}{\partial y} (5 x^{2} + 8 x y + y^{2}) = 2 y + 8 x

= \frac{1}{2 5 x ^{2} + 8 x y + y ^{2}} (2 y + 8 x)

Simplify

\frac{1}{2 5 x ^{2} + 8 x y + y ^{2}} (2 y + 8 x) : \frac{y + 4 x}{5 x ^{2} + 8 x y + y ^{2}}

= \frac{y + 4 x}{5 x ^{2} + 8 x y + y ^{2}}

\frac{d y}{d x} + \frac{5 x ^{2}}{e ^{3 x^{3}}} y = 0

x y^{^{'}} - 2 y = 2 x^{2} + x - 2

\int \frac{( x ^{2} - x + 4 )}{x ^{3} + 2 x} d x

\int \frac{( 1 - 2 x ) ^{2}}{x} d x

\frac{d}{d x} (- x^{2} + 25)

What is (\partial)/(\partial y)(sqrt(5x^2+8xy+y^2)) ?
The answer to (\partial)/(\partial y)(sqrt(5x^2+8xy+y^2)) is (y+4x)/(sqrt(5x^2+8xy+y^2))