What is the d/(du)(v/(u-v)) ?

The d/(du)(v/(u-v)) is -v/((u-v)^2)

What is the first d/(du)(v/(u-v)) ?

The first d/(du)(v/(u-v)) is -v/((u-v)^2)

Popular >

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

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

Solution steps

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

Treat

v

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d 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}}

\int_{0}^{2} x^{2} + 2 d x

\int \frac{- 4 ln ^{2} ( x )}{x} d x

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

\frac{\partial}{\partial v} (2 sin (u) sin (v))

\frac{d}{d x} (\frac{5 x}{x + 2})