What is (\partial)/(\partial u)(v/(1+uvw)) ?

The answer to (\partial)/(\partial u)(v/(1+uvw)) is -(v^2w)/((1+vuw)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial u)(v/(1+uvw))

\frac{\partial}{\partial u} (\frac{v}{1 + uv w})

- \frac{v ^{2} w}{( 1 + vu w ) ^{2}}

Solution steps

\frac{\partial}{\partial u} (\frac{v}{1 + uv w})

Treat

v, w

as constants

Take the constant out:

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

= v \frac{\partial}{\partial u} (\frac{1}{1 + uv w})

Apply exponent rule:

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

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

Apply the chain rule:

- \frac{1}{( 1 + uv w ) ^{2}} \frac{\partial}{\partial u} (1 + uv w)

= - \frac{1}{( 1 + uv w ) ^{2}} \frac{\partial}{\partial u} (1 + uv w)

\frac{\partial}{\partial u} (1 + uv w) = v w

= v (- \frac{1}{( 1 + uv w ) ^{2}} v w)

Simplify

v (- \frac{1}{( 1 + uv w ) ^{2}} v w) : - \frac{v ^{2} w}{( 1 + vu w ) ^{2}}

= - \frac{v ^{2} w}{( 1 + vu w ) ^{2}}

g (t) = - 4 t^{2} + 3 t - 3

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- t}

\int_{- 1}^{4} (x^{2} + x^{3}) d x

x^{3}

What is (\partial)/(\partial u)(v/(1+uvw)) ?
The answer to (\partial)/(\partial u)(v/(1+uvw)) is -(v^2w)/((1+vuw)^2)