What is (\partial)/(\partial x)((2x)/(1+x+2y)) ?

The answer to (\partial)/(\partial x)((2x)/(1+x+2y)) is (2(2y+1))/((1+x+2y)^2)

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(\partial)/(\partial x)((2x)/(1+x+2y))

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

\frac{\partial}{\partial x} (x) = 1

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

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

Simplify

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

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

f (x) = ln (2 x + 1)

\frac{e ^{x}}{8 x ^{2} + 1}

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

\int_{1}^{4} 5 - x d x

\int_{0}^{4} \frac{1}{64 - x ^{2}} d x

What is (\partial)/(\partial x)((2x)/(1+x+2y)) ?
The answer to (\partial)/(\partial x)((2x)/(1+x+2y)) is (2(2y+1))/((1+x+2y)^2)