What is (\partial)/(\partial x)(((2x+y))/(x+2y)) ?

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

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(\partial)/(\partial x)(((2x+y))/(x+2y))

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

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

Solution steps

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

Treat

y

as a constant

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

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

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

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

\int_{0}^{π} π (sin^{2} (x))^{2} d x

\int + x ln (x) d x

e^{4 x}, e^{8 x}, x = 1

(x^{2} + 1) y^{^{'}} + 6 x (y - 1) = 0

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