What is (\partial)/(\partial x)(y/(y-4x)) ?

The answer to (\partial)/(\partial x)(y/(y-4x)) is (4y)/((y-4x)^2)

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(\partial)/(\partial x)(y/(y-4x))

\frac{\partial}{\partial x} (\frac{y}{y - 4 x})

\frac{4 y}{( y - 4 x ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{y}{y - 4 x})

Treat

y

as a constant

Take the constant out:

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

= y \frac{\partial}{\partial x} (\frac{1}{y - 4 x})

Apply exponent rule:

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

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

Apply the chain rule:

- \frac{1}{( y - 4 x ) ^{2}} \frac{\partial}{\partial x} (y - 4 x)

= - \frac{1}{( y - 4 x ) ^{2}} \frac{\partial}{\partial x} (y - 4 x)

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

= y (- \frac{1}{( y - 4 x ) ^{2}} (- 4))

Simplify

y (- \frac{1}{( y - 4 x ) ^{2}} (- 4)) : \frac{4 y}{( y - 4 x ) ^{2}}

= \frac{4 y}{( y - 4 x ) ^{2}}

\frac{( e ^{x} )}{1 + x}

x \to \infty lim (\frac{x ^{4}}{e ^{- 6 x}})

y = 2 ln (x), y = 0, x = 3 e

\int_{1}^{- 2} (\frac{1}{2}) d x

\frac{5}{s ^{2} + 5}

What is (\partial)/(\partial x)(y/(y-4x)) ?
The answer to (\partial)/(\partial x)(y/(y-4x)) is (4y)/((y-4x)^2)