Find the following derivatives by using Chain Rule. a) dz/dt, where z = x sin y, x = t², and y = 4t³ [Ans: 2t sin 4t³ + 12t4 cos 4t³] b) dw/dt, where w = cos 2x sin 3y, x = t/2, and y = t*. [Ans: – sin t sin 3t* + 12t3 cos t cos 3t4] c) dw/dt, where w = xy sin z, x = t², y = 4t³, and z = t + 1. %3D [Ans: 20t4 sin(t + 1) + 4t5 cos(t + 1)] d) dU/dt, where U = In(x + y + z), x = t, y = t², and z = t³. %3D 1+2t+3t². [Ans: t+t2+t3

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Author:James Stewart
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Chapter1: Functions And Models
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Find the following derivatives by using Chain Rule.
a) dz/dt, where z = x sin y, x = t², and y = 4t3
[Ans: 2t sin 4t3 + 12t* cos 4t³]
COS
b) dw/dt, where w = cos 2x sin 3y, x = t/2, and y = t4.
[Ans: – sin t sin 3t4 + 12t³ cos t cos 3t4]
c) dw/dt, where w = xy sin z, x = t², y = 4t³, and z = t + 1.
[Ans: 20t4 sin(t + 1) + 4t5 cos(t + 1)]
d) dU/dt, where U = ln(x+ y + z), x = t, y = t², and z = t³.
1+2t+3t².
[Ans:
t+t2+t3
e) Zs and zt, where z = xy – x²y, x = s+ t, and y = s – t.
[Ans:z, = 2s – 3s² – 2st + t²; z, = -s² – 2t + 2st + 3t²]
where z = e*+y, x = st, and y = s+t.
f)
Zs
and
Zt,
[Ans:z, = (t + 1)est+s+t, z¢ = (s + 1)est+s+t]
x-z
g) ws and Wt, where w =
x = s+t, y = st and z = s – t
y+z'
-2t(t+1)
2s
[Ans:ws
; W¢ =
(st+s-t)2
(st+s-t)2-
Transcribed Image Text:Find the following derivatives by using Chain Rule. a) dz/dt, where z = x sin y, x = t², and y = 4t3 [Ans: 2t sin 4t3 + 12t* cos 4t³] COS b) dw/dt, where w = cos 2x sin 3y, x = t/2, and y = t4. [Ans: – sin t sin 3t4 + 12t³ cos t cos 3t4] c) dw/dt, where w = xy sin z, x = t², y = 4t³, and z = t + 1. [Ans: 20t4 sin(t + 1) + 4t5 cos(t + 1)] d) dU/dt, where U = ln(x+ y + z), x = t, y = t², and z = t³. 1+2t+3t². [Ans: t+t2+t3 e) Zs and zt, where z = xy – x²y, x = s+ t, and y = s – t. [Ans:z, = 2s – 3s² – 2st + t²; z, = -s² – 2t + 2st + 3t²] where z = e*+y, x = st, and y = s+t. f) Zs and Zt, [Ans:z, = (t + 1)est+s+t, z¢ = (s + 1)est+s+t] x-z g) ws and Wt, where w = x = s+t, y = st and z = s – t y+z' -2t(t+1) 2s [Ans:ws ; W¢ = (st+s-t)2 (st+s-t)2-
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