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richyw
 2 years ago
Best ResponseYou've already chosen the best response.0so I have \(k(x,y,z)=e^{x^2+y^2+z^2}\) where \(x=\sqrt{t+1},\;y=\sqrt{t^2+1},\:z=\sqrt{t^3+1}\) . Now i need to find \(\frac{dk}{dt}\).

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0to do this do I just say\[\frac{dk}{dt}=\frac{dk}{dx}\frac{dx}{dt}+\frac{dk}{dy}\frac{dy}{dt}+\frac{dk}{dz}\frac{dz}{dt}\]

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0I am getting an answer of \[\frac{dk}{dt}=2e^{t^3+t^2+t+3}\left(3t^2+2t+1\right)\] is this correct, and does anyone know how to check this on wolfram, or opensource math software?

eseidl
 2 years ago
Best ResponseYou've already chosen the best response.1The formula for the total differential is correct, as long as the dk/dx,etc terms are partial derivatives.

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0yeah sorry I didn't want to type "partial" \(\partial\) in the latex that many times :)

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0thanks a lot. can anyone confirm this is the correct answer?

eseidl
 2 years ago
Best ResponseYou've already chosen the best response.1this one is super tedious...give me a minute I'll do it with pen and paper...

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0alright thanks a lot. I really appreciate it.

eseidl
 2 years ago
Best ResponseYou've already chosen the best response.1I get:\[e^{t^3+t^2+t+3}(3t^2+2t+1)\]

eseidl
 2 years ago
Best ResponseYou've already chosen the best response.1Your factor of two should cancel when you differentiate the square root term (which gives you a 2 in the denominator)...

richyw
 2 years ago
Best ResponseYou've already chosen the best response.0thanks a lot! I see I made a basic calculus mistake but am glad I remember this stuff!
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