What is the derivative of e^{xy}+ln(x/y) ?

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

What is the first derivative of e^{xy}+ln(x/y) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (e^{x y} + ln (\frac{x}{y}))

e^{x y} (y + x \frac{d y}{d x}) + \frac{y - x \frac{d y}{d x}}{x y}

Solution steps

\frac{d}{d x} (e^{x y} + ln (\frac{x}{y}))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (e^{x y}) + \frac{d}{d x} (ln (\frac{x}{y}))

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

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

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

\frac{d y}{d x} = 4 + y - 4 x + 2

x \to 0 lim (a x)

\frac{d}{d x} (2 x^{2 x})

\frac{\partial}{\partial x} (x - y^{4})

x \to 0 lim (\frac{tan ( a ) x}{x})

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