What is (\partial)/(\partial u)(v/(u-v)) ?

The answer to (\partial)/(\partial u)(v/(u-v)) is -v/((u-v)^2)

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(\partial)/(\partial u)(v/(u-v))

\frac{\partial}{\partial u} (\frac{v}{u - v})

- \frac{v}{( u - v ) ^{2}}

Solution steps

\frac{\partial}{\partial u} (\frac{v}{u - v})

Treat

v

as a constant

Take the constant out:

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

= v \frac{\partial}{\partial u} (\frac{1}{u - v})

Apply exponent rule:

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

= v \frac{\partial}{\partial u} ((u - v)^{- 1})

Apply the chain rule:

- \frac{1}{( u - v ) ^{2}} \frac{\partial}{\partial u} (u - v)

= - \frac{1}{( u - v ) ^{2}} \frac{\partial}{\partial u} (u - v)

\frac{\partial}{\partial u} (u - v) = 1

= v (- \frac{1}{( u - v ) ^{2}} \cdot 1)

Simplify

v (- \frac{1}{( u - v ) ^{2}} \cdot 1) : - \frac{v}{( u - v ) ^{2}}

= - \frac{v}{( u - v ) ^{2}}

\frac{d y}{d x} = 2 sin (x) (y - 3)

\frac{d}{d x} (\frac{3}{2 3 x + 1})

x \to 0 lim (4 cos (2 x))

y^{^{''}} - y^{^{'}} = 6 x e^{x}

x \to 0 lim (x^{3} - 3 x^{2} + 2 x - 3)

What is (\partial)/(\partial u)(v/(u-v)) ?
The answer to (\partial)/(\partial u)(v/(u-v)) is -v/((u-v)^2)