What is the partial derivative of u with respect to t in terms of x, y, and t?

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SUMMARY

The discussion focuses on calculating the partial derivative of the function u = e^{-xy} with respect to time t, given the parametric equations x = cosh(t) and y = sinh(t). The user successfully derives the partial derivatives ∂u/∂x = -ye^{-xy} and ∂u/∂y = -xe^{-xy}. By applying the chain rule, the user concludes that du/dt = ∂u/∂x * dx/dt + ∂u/∂y * dy/dt, which simplifies the process of finding du/dt in terms of x, y, and t.

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jamesbob
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This is annoying me as i have the answer on the tip of my pen, just can't write it down. I'm not 100% sure i understand what the question is asking me to do.

Consider the quantity [tex]u = e^{-xy}[/tex] where (x,y) moves in time t along a path:

[tex]x = \cosh{t}, \mbox{ } y = \sinh{t}[/tex]​

Use a method based on partial derivatives to calculate [tex]\frac{du}{dt}[/tex] as a function of x, y and t.

My answer:

I partially differentiated u, getting:

[tex]\frac{\delta{u}}{\delta{x}} = -ye^{-xy}[/tex]
[tex]\frac{\delta{u}}{\delta{y}} = -xe^{-xy}[/tex]
So does this mean [tex]du = -ye^{-xy} + -xe^{-xy} ?[/tex]

I though that i would get du from the part iv just explained, then get dt from differentiating x and y. But this ofcourse leaves me with expressions for dx/dt and dy/dt. Where do i go from here?
 
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Use the chain rule:
[tex]\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt}[/tex]
 
Last edited:
Yeah, realized that after some research - just something id never saw. Dead easy tho. Thanks anyway :smile:
 

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