What is (\partial)/(\partial y)((3x)/((x+y)^4)) ?

The answer to (\partial)/(\partial y)((3x)/((x+y)^4)) is -(12x)/((x+y)^5)

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(\partial)/(\partial y)((3x)/((x+y)^4))

\frac{\partial}{\partial y} (\frac{3 x}{( x + y ) ^{4}})

- \frac{12 x}{( x + y ) ^{5}}

Solution steps

\frac{\partial}{\partial y} (\frac{3 x}{( x + y ) ^{4}})

Treat

x

as a constant

Take the constant out:

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

= 3 x \frac{\partial}{\partial y} (\frac{1}{( x + y ) ^{4}})

Apply exponent rule:

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

= 3 x \frac{\partial}{\partial y} ((x + y)^{- 4})

Apply the chain rule:

- \frac{4}{( x + y ) ^{5}} \frac{\partial}{\partial y} (x + y)

= - \frac{4}{( x + y ) ^{5}} \frac{\partial}{\partial y} (x + y)

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

= 3 x (- \frac{4}{( x + y ) ^{5}} \cdot 1)

Simplify

3 x (- \frac{4}{( x + y ) ^{5}} \cdot 1) : - \frac{12 x}{( x + y ) ^{5}}

= - \frac{12 x}{( x + y ) ^{5}}

\int 3 x^{2} - 3 d x

f^{^{'}} (x) = 3 e^{x} + 7 sec^{2} (x)

2 y^{^{''}} + y^{^{'}} - y = x + 5

(x + y)^{2} d x + (2 x y + x^{2} - 8) d y = 0, y (1) = 1

\frac{d}{d x} (ln (\frac{x - 2}{x + 2}))

What is (\partial)/(\partial y)((3x)/((x+y)^4)) ?
The answer to (\partial)/(\partial y)((3x)/((x+y)^4)) is -(12x)/((x+y)^5)