What is (\partial)/(\partial x)((6y)/(x+y)) ?

The answer to (\partial)/(\partial x)((6y)/(x+y)) is -(6y)/((x+y)^2)

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(\partial)/(\partial x)((6y)/(x+y))

\frac{\partial}{\partial x} (\frac{6 y}{x + y})

- \frac{6 y}{( x + y ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{6 y}{x + y})

Treat

y

as a constant

Take the constant out:

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

= 6 y \frac{\partial}{\partial x} (\frac{1}{x + y})

Apply exponent rule:

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

= 6 y \frac{\partial}{\partial x} ((x + y)^{- 1})

Apply the chain rule:

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

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

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

= 6 y (- \frac{1}{( x + y ) ^{2}} \cdot 1)

Simplify

6 y (- \frac{1}{( x + y ) ^{2}} \cdot 1) : - \frac{6 y}{( x + y ) ^{2}}

= - \frac{6 y}{( x + y ) ^{2}}

\int e^{x} \cdot (1 - e^{x})^{- 1} d x

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

\frac{3}{x}

f (x) = x^{3} - 2 x + 1

- 3 sin (2 t)

What is (\partial)/(\partial x)((6y)/(x+y)) ?
The answer to (\partial)/(\partial x)((6y)/(x+y)) is -(6y)/((x+y)^2)