Question
Show that $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$ for $u=e^{x} \cos y$and $v=e^{x} \sin y$
Step 1
For $u=e^{x} \cos y$, the partial derivative with respect to $x$ is obtained by differentiating $e^{x}$ with respect to $x$ and keeping $y$ constant. This gives us: \[\frac{\partial u}{\partial x}=e^{x} \cos y\] Show more…
Show all steps
Your feedback will help us improve your experience
Adrian Co and 77 other Calculus 3 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Show that $\frac{\partial w}{\partial u}+\frac{\partial w}{\partial v}=0$ for $w=f(x, y), x=u-v,$ and $y=v-u$
Functions of Several Variables
Chain Rules for Functions of Several Variables
If $u=f(x, y),$ where $x=e^{x} \cos t$ and $y=e^{x} \sin t,$ show that $$ \left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}=e^{-2} \cdot\left[\left(\frac{\partial u}{\partial s}\right)^{2}+\left(\frac{\partial u}{\partial t}\right)^{2}\right] $$
Partial Derivatives
The Chain Rule
Show that $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$ for $u=\ln \sqrt{x^{2}+y^{2}}$ and $v=\tan ^{-1} \frac{y}{x}$
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD