What is (\partial)/(\partial y)(x/((1+2x+3y))) ?

The answer to (\partial)/(\partial y)(x/((1+2x+3y))) is -(3x)/((1+2x+3y)^2)

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(\partial)/(\partial y)(x/((1+2x+3y)))

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

2 y^{^{''}} - 9 y^{^{'}} + 4 y = 0

f (x) = x^{6} e^{x}

y = e^{- 2 x} \cdot tan (8 x)

x \to 0 lim (\frac{3 sin ( 3 x )}{x})

x, sin (x), [\frac{π}{2}, π]

What is (\partial)/(\partial y)(x/((1+2x+3y))) ?
The answer to (\partial)/(\partial y)(x/((1+2x+3y))) is -(3x)/((1+2x+3y)^2)