Recent content by inflectious
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I seem to lack an implicit differentiation technique, should be a quick fix
##y## isn't a constant with respect to ##x##. The original equation defines a function implicitly; the value of any parameter is dependent upon the other two. I've plugged the equation into mathematica, and the ##\frac{\partial y}{\partial x}## and ##\frac{\partial x}{\partial y}## terms which...- inflectious
- Post #6
- Forum: Calculus and Beyond Homework Help
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I
I seem to lack an implicit differentiation technique, should be a quick fix
\begin{cases} \frac{\partial u}{\partial x}=\frac{u}{x(1-u\ln{x})}+\frac{u\ln{u}}{1-y}\frac{\partial y}{\partial x}\\ \frac{\partial u}{\partial y}=\frac{u\ln{x}}{1-y}+\frac{u^2}{x(1-u\ln{x})}\frac{\partial x}{\partial y} \end{cases} is what I've got so far. I began doing a bunch of algebra...- inflectious
- Post #4
- Forum: Calculus and Beyond Homework Help
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I
I seem to lack an implicit differentiation technique, should be a quick fix
Thanks, Ayre! Seems to boil down to a combination of the ubiquitous chain rule for logarithms of functions and the product rule. So clear now! I should've seen it... hopefully a bit of mathematical intuition returns with study.- inflectious
- Post #3
- Forum: Calculus and Beyond Homework Help
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I
I seem to lack an implicit differentiation technique, should be a quick fix
Homework Statement u = x^u + u^y Find the partial derivatives of ##u## w.r.t. ##x, y##.Homework Equations Only the one. The Attempt at a Solution I've attempted reducing the problem using logs, but the resulting equations seem no more tenable to me. I'm sure there is a nice trick... it...- inflectious
- Thread
- Differentiation Implicit Implicit differentiation
- Replies: 8
- Forum: Calculus and Beyond Homework Help