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

The answer to (\partial)/(\partial y)((2x)/((2x+3y)^2)) is -(12x)/((2x+3y)^3)

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

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

- \frac{12 x}{( 2 x + 3 y ) ^{3}}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

2 x (- \frac{2}{( 2 x + 3 y ) ^{3}} \cdot 3) : - \frac{12 x}{( 2 x + 3 y ) ^{3}}

= - \frac{12 x}{( 2 x + 3 y ) ^{3}}

f (x) = e^{t}

ln ∣ sin (x) ∣

y = 2 x + 1, (4, 3)

2 t^{- \frac{1}{4}}

\int \frac{x ^{2}}{e ^{\frac{x}{3}}} d x

What is (\partial)/(\partial y)((2x)/((2x+3y)^2)) ?
The answer to (\partial)/(\partial y)((2x)/((2x+3y)^2)) is -(12x)/((2x+3y)^3)