Question

Use the Chain Rule to find dw/dt. (Enter your answer only in terms of t.)
w = xey /z, x = t6, y = 9 - t, z = 4 + 4t

Answer 1

Do you mean find dw/dt where\[w=\frac{xe^y}{z}, x=t^6, y=9-t, z=4+4t?\]

Answer 2

yes

Answer 3

i got an answer I keep on getting the same one, but my hw is online and it marks it wrong

Answer 4

OK. If you really *must* use the chain rule to solve,\[\frac{dw}{dt}=\frac{dw}{dx}\frac{dx}{dy}\frac{dy}{dz}\frac{dz}{dt}\]Then expand dx/dy and dy/dz again\[\frac{dw}{dt}=\frac{dw}{dx}(\frac{dx}{dt}\frac{dt}{dy})(\frac{dy}{dt}\frac{dt}{dz})\frac{dz}{dt}\]

Answer 5

this is for calc 3 so I have to multiply the leibniz notation and add them

Answer 6

Ah, okay, knowing the level helps.

Answer 7

Well, in that case, it's just\[\frac{dw}{dt}=\frac{dw}{dx}\frac{dx}{dt}+\frac{dw}{dy}\frac{dy}{dt}+\frac{dw}{dz}\frac{dz}{dt}\] where all the 'd's are partial.

Answer 8

yes I have computed that

Answer 9

I'm doing the rest...

Answer 10

hey lokisan do you get the divisible sign to be straight instead of this /

Answer 11

I get \[\frac{dw}{dt}=\frac{-4t^6}{(4+4t)^2}e^{9-t}\]

Answer 12

Did you need just a check on your answer, or the full-on working?

Answer 13

nadeem, you type "frac{}{}" into the equation editor. Your numerator and denominator go in the first and second parentheses respectively.

Answer 14

that's wassup..... appreciate it

Answer 15

no probs

Answer 16

is that what you get? how did you get that?

Answer 17

Just doing what each derivative asks of me first\[\frac{dw}{dt}=(\frac{e^y}{z}).t^6+\frac{xe^y}{z}.(-1)+(-\frac{xe^y}{z^2}).4\]

Answer 18

Then substitute each of the x=t^6, y=... into the above. You should find that the first two quotients cancel out, and you're left with the last quotient -> the answer.

Answer 19

This is what I got:
\[\frac{dw}{dt}=\frac{-4t^5e^{9-t}(t^2-4t-6)}{(4+4t)^2}\]

Answer 20

\[\frac{dw}{dt}=(\frac{e^{9-t}}{4+4t}).t^6-\frac{t^6e^{9-t}}{4+4t}-4\frac{t^6e^{9-t}}{(4+4t)^2}\]

Answer 21

The first two parts cancel, the third is left over.