What is (\partial)/(\partial x)(-(2y)/((2x+3y)^2)) ?

The answer to (\partial)/(\partial x)(-(2y)/((2x+3y)^2)) is (8y)/((2x+3y)^3)

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(\partial)/(\partial x)(-(2y)/((2x+3y)^2))

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

\frac{8 y}{( 2 x + 3 y ) ^{3}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= - 2 y \frac{\partial}{\partial x} (\frac{1}{( 2 x + 3 y ) ^{2}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

- 2 y (- \frac{2}{( 2 x + 3 y ) ^{3}} \cdot 2) : \frac{8 y}{( 2 x + 3 y ) ^{3}}

= \frac{8 y}{( 2 x + 3 y ) ^{3}}

y^{^{'}} = 2 y, y (0) = 1

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

x \to 0 lim (\frac{5}{cos ( 5 x )})

\frac{\partial}{\partial x} (3 x y - x^{2} y)

\int_{2}^{4} (\frac{1}{2} x^{3} - 3 x + 2) d x

What is (\partial)/(\partial x)(-(2y)/((2x+3y)^2)) ?
The answer to (\partial)/(\partial x)(-(2y)/((2x+3y)^2)) is (8y)/((2x+3y)^3)