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

The answer to (\partial)/(\partial x)(3/(3x+y)) is -9/((3x+y)^2)

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

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

- \frac{9}{( 3 x + y ) ^{2}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

3 (- \frac{1}{( 3 x + y ) ^{2}} \cdot 3) : - \frac{9}{( 3 x + y ) ^{2}}

= - \frac{9}{( 3 x + y ) ^{2}}

\int - \frac{1}{2} y d y

x \to - 1 lim (x + 3)

\frac{0.6}{s + 2}

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

\int e^{u} d u

What is (\partial)/(\partial x)(3/(3x+y)) ?
The answer to (\partial)/(\partial x)(3/(3x+y)) is -9/((3x+y)^2)