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stamp
 3 years ago
[CALCULUS III—CHAIN RULE] Find dw/dt by the chain rule where w = xy cos(z) and x = t, y = t^2, and z = arccos(t).
stamp
 3 years ago
[CALCULUS III—CHAIN RULE] Find dw/dt by the chain rule where w = xy cos(z) and x = t, y = t^2, and z = arccos(t).

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stamp
 3 years ago
Best ResponseYou've already chosen the best response.0\[w=xy\ cos(z)\]\[w=t(t^2)\ cos(arccos(t))\]\[w=t^3t\]\[w=t^4\]\[dw/dt=4t^3\] Is that all there is to it or am I mistaken?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1by the chain rule eh \[w=xy\ cos(z)\] \[w(t)=x(t)y(t)\ cos(z(t))\] \[w'(t)=x'(t)y(t)\ cos(z(t))+x(t)y'(t)\ cos(z(t))+x(t)y(t)\ cos'(z(t))\] \[w'(t)=x'(t)y(t)\ cos(z(t))+x(t)y'(t)\ cos(z(t))x(t)y(t)z'(t)\ sin(z(t))\]

stamp
 3 years ago
Best ResponseYou've already chosen the best response.0Or would all that simplify to 4t^3

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1we can ignore the (t) parts to clean it up \[w'=x'y\ cos(z)+xy'\ cos(z)xyz'\ sin(z)\] \[w'=cos(z)(x'y+xy')xyz'\ sin(z)\] \[w'=cos(z)(x'y+xy')xyz'\ sin(z)\] x = t, y = t^2, and z = arccos(t) x = 1, y = 2t, and z = \(\frac{1}{\sqrt{1+...}}\) \[w'=cos(cos^{1}z)(y+2t^2)t^3z'\ sin(cos^{1}z)\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1\[w'=z(t^2+2t^2)t^3z'\ sin(cos^{1}z)\] \[w'=3t^2zt^3z'\ sin(cos^{1}z)\] is that looking right so far?

stamp
 3 years ago
Best ResponseYou've already chosen the best response.0You forgot your x' and y' and z' on your derivatives but it is ok because I recognize what you did.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1it does simplify out to 4t^3

stamp
 3 years ago
Best ResponseYou've already chosen the best response.0\[w′=cos(z)(x′y+xy′)−xyz′ sin(z)\] x'y = t^2 xy' = 2t x'y + xy' = 2t + t^2 = t(2+t) cos(z) = cos(arccos t) = t t( t(2+t) )  xyz' sin(z)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1provided t not equal 1 or 1 due to that z'

stamp
 3 years ago
Best ResponseYou've already chosen the best response.0I am not quite seeing the simplification to 4t^3

stamp
 3 years ago
Best ResponseYou've already chosen the best response.0I understand the first part where you showed the formal chain rule of w(t) = x(t) y(t) cos(z(t)) The simplification of it to 4t^3 is still eluding me.

stamp
 3 years ago
Best ResponseYou've already chosen the best response.0ps thank you for your assistance so far

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1i got lost, let me go back to the w' chains :) and sort it back out

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1\[w'=x'y\cos(z)+xy'\cos(z)+xy~\cos'(z(t))\] \[w'=x'y\cos(z)+xy'\cos(z)xyz'~\sin(z)\] \[w'=x'y\cos(z)+xy'\cos(z)xyz'~\sin(z)\] \[x = t,~ y = t^2, ~z = cos^{1}(t)\]\[x' = 1, ~y' = 2t, ~z' =(1t^2)^{(1/2)}\] \[w'=t^2t+t~2t~ttt^2(1t^2)^{(1/2)}~\sin(\cos^{1}(t))\] \[w'=t^3+2t^3t^3(1t^2)^{(1/2)}~\sin(\cos^{1}(t))\] \[w'=t^3(3(1t^2)^{(1/2)}~\sin(\cos^{1}(t))\] the key here is in definine sin(cos^1 (t)) better

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1361983895329:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1we still have to include the caveat of t not equal to 1 or 1

stamp
 3 years ago
Best ResponseYou've already chosen the best response.0sin(arccos(t)) = sqrt( 1  t^2 )

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1so that reduces to w' = (31) t^3 which means i prolly missed a sign in there someplace along the way :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1i forgot the  part :)

stamp
 3 years ago
Best ResponseYou've already chosen the best response.0w' = (31)t^3 w' = 2t^3 ?? :(

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1check out z', its spose to be negative ... not positive

stamp
 3 years ago
Best ResponseYou've already chosen the best response.0Ok. I will ponder all of this and do some work on paper to verify everything. Thank you again for your time.

stamp
 3 years ago
Best ResponseYou've already chosen the best response.0@amistre64 Here is the derivative of x(t) y(t) cos z(t) http://www.wolframalpha.com/input/?i=derivative+of+x%28t%29y%28t%29+cos%28z%28t%29%29+

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1yes, but z' is negative so that changes the sign .... and carries it to the end as (3+1) not (31)
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