What is the derivative of ln(yx) ?

The derivative of ln(yx) is (x\frac{dy)/(dx)+y}{xy}

What is the first derivative of ln(yx) ?

The first derivative of ln(yx) is (x\frac{dy)/(dx)+y}{xy}

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\frac{d}{d x} (ln (y x))

\frac{x \frac{d y}{d x} + y}{x y}

Solution steps

\frac{d}{d x} (ln (y x))

Apply the chain rule:

\frac{1}{y x} \frac{d}{d x} (y x)

= \frac{1}{y x} \frac{d}{d x} (y x)

\frac{d}{d x} (y x) = x \frac{d y}{d x} + y

= \frac{1}{y x} (x \frac{d y}{d x} + y)

Simplify

\frac{1}{y x} (x \frac{d y}{d x} + y) : \frac{x \frac{d y}{d x} + y}{x y}

= \frac{x \frac{d y}{d x} + y}{x y}

x y^{^{'}} - (1 + x) y = x y^{2}

x \to 1 lim (\frac{x - 1}{x + 24 - 5})

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

n = 1 \sum \infty \frac{n ^{4}}{n ^{5}}

\frac{\partial}{\partial z} (2 x - 3 yz)

What is the derivative of ln(yx) ?
The derivative of ln(yx) is (x\frac{dy)/(dx)+y}{xy}
What is the first derivative of ln(yx) ?
The first derivative of ln(yx) is (x\frac{dy)/(dx)+y}{xy}