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)

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

Simplify

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

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

\int (4 x^{3} - 9 x^{2} + 16 x) x^{4} - 3 x^{3} + 8 x^{2} d x

\frac{\partial}{\partial x} (e^{5 x} cos (5 y))

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

x \to 0 lim (\frac{4 + x - 4 - x}{4 x})

x \to - \frac{π}{2} lim (- tan (0.5 x))

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)