What is (\partial)/(\partial x)(((7x-8y))/(1x+2y)) ?

The answer to (\partial)/(\partial x)(((7x-8y))/(1x+2y)) is (22y)/((x+2y)^2)

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(\partial)/(\partial x)(((7x-8y))/(1x+2y))

\frac{\partial}{\partial x} (\frac{( 7 x - 8 y )}{1 x + 2 y})

\frac{22 y}{( x + 2 y ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{( 7 x - 8 y )}{1 \cdot 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} ( 7 x - 8 y ) ( 1 \cdot x + 2 y ) - \frac{\partial}{\partial x} ( 1 \cdot x + 2 y ) ( 7 x - 8 y )}{( 1 \cdot x + 2 y ) ^{2}}

\frac{\partial}{\partial x} (7 x - 8 y) = 7

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

= \frac{7 ( 1 \cdot x + 2 y ) - 1 \cdot ( 7 x - 8 y )}{( 1 \cdot x + 2 y ) ^{2}}

Simplify

\frac{7 ( 1 \cdot x + 2 y ) - 1 \cdot ( 7 x - 8 y )}{( 1 \cdot x + 2 y ) ^{2}} : \frac{22 y}{( x + 2 y ) ^{2}}

= \frac{22 y}{( x + 2 y ) ^{2}}

\frac{d}{d x} (\frac{cos ( x ^{2} )}{x ^{2} + 1})

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

\frac{\partial}{\partial x} (x^{0.5} y^{0.5} z)

(x + 2)^{2} y^{^{'}} + 8 y = 5 - 4 x y

x \to 4 lim (x^{2} - 5 x + 7 x^{3})

What is (\partial)/(\partial x)(((7x-8y))/(1x+2y)) ?
The answer to (\partial)/(\partial x)(((7x-8y))/(1x+2y)) is (22y)/((x+2y)^2)