Question

Use the Chain Rule to evaluate the partial derivative ∂g/∂u at the point (u,v)=(0,1), where g(x,y)=x^2−y^2, x=e^3ucos(v), y=e^3usin(v).

(Use symbolic notation and fractions where needed.)

Answer #1

Use the Chain Rule to evaluate the partial derivative
∂f∂u and ∂f∂u at (u, v)=(−1, −1), where
f(x, y, z)=x10+yz16,
x=u2+v, y=u+v2, z=uv.
(Give your answer as a whole or exact number.)
∂f∂u=
∂f∂v=

Let f(x,y,z)=6xy−z^2, x=6rcos(θ), y=cos^2(θ), z=7r.
Use the Chain Rule to calculate the partial derivative.
(Use symbolic notation and fractions where needed. Express the
answer in terms of independent variables

Let f(x,y,z)=xy+z^3, x=r+s−8t, y=3rt, z=s^6.
Use the Chain Rule to calculate the partial derivatives.
(Use symbolic notation and fractions where needed. Express the
answer in terms of independent variables

For the function w=f(x,y) , x=g(u,v) , and
y=h(u,v). Use the Chain Rule to
Find ∂w/∂u and
∂w/∂v when u=2 and v=3 if
g(2,3)=4, h(2,3)=-2,
gu(2,3)=-5,
gv(2,3)=-1 ,
hu(2,3)=3,
hv(2,3)=-5,
fx(4,-2)=-4, and
fy(4,-2)=7
∂w/∂u=
∂w/∂v =

1.
a) Use the Chain Rule to calculate the partial derivatives.
Express the answer in terms of the independent variables.
∂f
∂r
∂f
∂t
; f(x, y, z) = xy +
z2, x = r + s −
2t, y = 6rt, z =
s2
∂f
∂r
=
∂f
∂t
=
b) Use the Chain Rule to calculate the partial derivative.
Express the answer in terms of the independent variables.
∂F
∂y
; F(u, v) =
eu+v, u =
x5, v = 2xy
∂F
∂y
=
c)...

if y=uv, where u and v are functions of x, show that the nth
derivative of y with respect to x is given by
(also known as Leibniz Rule)

Use the Chain Rule to find the indicated partial
derivatives.
z = x2 + xy3,
x = uv2 + w3,
y = u + vew
when u = 2, v = 2, w = 0

Use the Chain Rule to find the indicated partial derivatives. u
= x^4 + yz, x = pr sin(θ), y = pr cos(θ), z = p + r; (partial
u)/(partial p), (partial u)/(partial r), (partial u)/(partial
theta) when p = 3, r = 4, θ = 0

Use the Chain Rule to find the indicated partial derivatives. u
=sqrt( r^2 + s^2) , r = y + x cos(t), s = x + y sin(t)
∂u ∂x , ∂u ∂y , ∂u ∂t when x = 1, y = 4, t = 0

Find the area enclosed by y=ln(x−8) and y=(ln(x−8))^(2).
(Use symbolic notation and fractions where needed.)
A=

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