What is (\partial)/(\partial x)(-2kl(ln(y/(-x-y)))) ?

The answer to (\partial)/(\partial x)(-2kl(ln(y/(-x-y)))) is -(2lk)/(-x-y)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(-2kl(ln(y/(-x-y))))

\frac{\partial}{\partial x} (- 2 k l (ln (\frac{y}{- x - y})))

- \frac{2 l k}{- x - y}

Solution steps

\frac{\partial}{\partial x} (- 2 k l (ln (\frac{y}{- x - y})))

Treat

l, k, y

as a constant

Take the constant out:

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

= - 2 k l \frac{\partial}{\partial x} (ln (\frac{y}{- x - y}))

Apply the chain rule:

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

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

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

= - 2 k l \frac{1}{\frac{y}{- x - y}} \cdot \frac{y}{( - x - y ) ^{2}}

Simplify

- 2 k l \frac{1}{\frac{y}{- x - y}} \cdot \frac{y}{( - x - y ) ^{2}} : - \frac{2 l k}{- x - y}

= - \frac{2 l k}{- x - y}

\int \frac{1}{6 - x} d x

g (x) = 2 x - 3

\int 14 θ sec^{2} (θ) d θ

\frac{d}{d x} (9 - x^{2} - y^{2})

3 (x^{3} - x^{2})

What is (\partial)/(\partial x)(-2kl(ln(y/(-x-y)))) ?
The answer to (\partial)/(\partial x)(-2kl(ln(y/(-x-y)))) is -(2lk)/(-x-y)