What is (\partial)/(\partial x)(ln(x-e^xy)) ?

The answer to (\partial)/(\partial x)(ln(x-e^xy)) is (1-e^xy)/(x-e^xy)

Popular >

(\partial)/(\partial x)(ln(x-e^xy))

\frac{\partial}{\partial x} (ln (x - e^{x} y))

\frac{1 - e ^{x} y}{x - e ^{x} y}

Solution steps

\frac{\partial}{\partial x} (ln (x - e^{x} y))

Treat

y

as a constant

Apply the chain rule:

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

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

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

= \frac{1}{x - e ^{x} y} (1 - y e^{x})

Simplify

\frac{1}{x - e ^{x} y} (1 - y e^{x}) : \frac{1 - e ^{x} y}{x - e ^{x} y}

= \frac{1 - e ^{x} y}{x - e ^{x} y}

y^{^{''}} + 16 y = 8 sin (4 t)

\int \frac{1}{x ^{2} + 4 x + 9} d x

y (t + 5) (t + 2) y^{^{'}} + t + 3 = 0

y = (x - 10)^{- 3}

f (x) = \frac{x}{x + 4}

What is (\partial)/(\partial x)(ln(x-e^xy)) ?
The answer to (\partial)/(\partial x)(ln(x-e^xy)) is (1-e^xy)/(x-e^xy)