Question
Find the following derivatives: (i)
z
s
and
z
t
where
z=e
x+y
,x=st
and
y=s+t
(ii)
z
s
and
z
t
where
z=sin(2x+y),x=s
2
−t
2
and
y=s
2
+t
2
(iii)
w
s
and
w
t
where
w=
y+z
x−z
,x=s+t
and
y=st
and
z=s−t
z
s
and
z
t
where
z=e
x+y
,x=st
and
y=s+t
(ii)
z
s
and
z
t
where
z=sin(2x+y),x=s
2
−t
2
and
y=s
2
+t
2
(iii)
w
s
and
w
t
where
w=
y+z
x−z
,x=s+t
and
y=st
and
z=s−t
Answers
GPT 3.5
(i) To find the derivatives zₛ, and zₜ, we need to use the chain rule.
First, let's find the derivative of z with respect to x:
dz/dx = d(e^(x+y))/dx
= e^(x+y) * (1 + dy/dx) (by chain rule)
= e^(x+y) * (1 + dy/dt * dt/dx) (since x = st and y = s + t)
= e^(x+y) * (1 + s*dt/dt)
= e^(x+y) * (1 + s) (since dt/dt = 1)
Next, let's find the derivative of z with respect to y:
dz/dy = d(e^(x+y))/dy
= e^(x+y) * (1 + dx/dy)
= e^(x+y) * (1 + dt/dy * dy/dx) (since x = st and y = s + t)
= e^(x+y) * (1 + t*dt/dt)
= e^(x+y) * (1 + t) (since dt/dt = 1)
Now, let's find the derivative of z with respect to s:
zₛ = dz/ds
= dz/dx * dx/ds + dz/dy * dy/ds
= e^(x+y) * (1 + s) * dt/ds + e^(x+y) * (1 + t) * dt/ds
= e^(x+y) * (1 + s + t) * dt/ds
Finally, let's find the derivative of z with respect to t:
zₜ = dz/dt
= dz/dx * dx/dt + dz/dy * dy/dt
= e^(x+y) * (1 + s) * ds/dt + e^(x+y) * (1 + t) * ds/dt
= e^(x+y) * (s + t + st) * ds/dt
(ii) To find the derivatives zₛ, and zₜ, we need to use the chain rule.
First, let's find the derivatives of x and y with respect to s and t:
dx/ds = d(s^2 - t^2)/ds
= 2s
dx/dt = d(s^2 - t^2)/dt
= -2t
dy/ds = d(s^2 + t^2)/ds
= 2s
dy/dt = d(s^2 + t^2)/dt
= 2t
Next, let's find the derivatives of z with respect to x and y:
dz/dx = d(sin(2x+y))/dx
= 2cos(2x+y) * dx/dx
= 2cos(2x+y)
dz/dy = d(sin(2x+y))/dy
= cos(2x+y) * dy/dy
= cos(2x+y)
Now, let's find the derivatives of z with respect to s and t:
zₛ = dz/ds
= dz/dx * dx/ds + dz/dy * dy/ds
= 2cos(2x+y) * 2s + cos(2x+y) * 2s
= 4s * cos(2x+y)
zₜ = dz/dt
= dz/dx * dx/dt + dz/dy * dy/dt
= 2cos(2x+y) * (-2t) + cos(2x+y) * 2t
= -4t * cos(2x+y)
(iii) To find the derivatives wₛ and wₜ, we need to use the quotient rule.
First, let's find the derivatives of x, y, and z with respect to s and t:
dx/ds = d(s + t)/ds
= 1
dx/dt = d(s + t)/dt
= 1
dy/ds = d(st)/ds
= t
dy/dt = d(st)/dt
= s
dz/ds = d(s - t)/ds
= 1
dz/dt = d(s - t)/dt
= -1
Now, let's find the derivatives of w with respect to s and t:
wₛ = [dy/ds * (x - z) - y * dz/ds] / (x - z)^2
= [t * (s + t - (s - t)) - st * 1] / (s + t - (s - t))^2
= [2t^2 - st] / (2t)^2
= (2t^2 - st) / 4t^2
= (t(2t - s)) / 4t^2
= (2t - s) / 4t
wₜ = [dy/dt * (x - z) - y * dz/dt] / (x - z)^2
= [s * (s + t - (s - t)) - st * (-1)] / (s + t - (s - t))^2
= [s^2 + st + st] / (2s)^2
= (2st + s^2) / 4s^2
= (s + t) / (2s)
Note: In both cases (i.e., wₛ and wₜ), the expressions can be further simplified, but this is the general form of the derivative.
First, let's find the derivative of z with respect to x:
dz/dx = d(e^(x+y))/dx
= e^(x+y) * (1 + dy/dx) (by chain rule)
= e^(x+y) * (1 + dy/dt * dt/dx) (since x = st and y = s + t)
= e^(x+y) * (1 + s*dt/dt)
= e^(x+y) * (1 + s) (since dt/dt = 1)
Next, let's find the derivative of z with respect to y:
dz/dy = d(e^(x+y))/dy
= e^(x+y) * (1 + dx/dy)
= e^(x+y) * (1 + dt/dy * dy/dx) (since x = st and y = s + t)
= e^(x+y) * (1 + t*dt/dt)
= e^(x+y) * (1 + t) (since dt/dt = 1)
Now, let's find the derivative of z with respect to s:
zₛ = dz/ds
= dz/dx * dx/ds + dz/dy * dy/ds
= e^(x+y) * (1 + s) * dt/ds + e^(x+y) * (1 + t) * dt/ds
= e^(x+y) * (1 + s + t) * dt/ds
Finally, let's find the derivative of z with respect to t:
zₜ = dz/dt
= dz/dx * dx/dt + dz/dy * dy/dt
= e^(x+y) * (1 + s) * ds/dt + e^(x+y) * (1 + t) * ds/dt
= e^(x+y) * (s + t + st) * ds/dt
(ii) To find the derivatives zₛ, and zₜ, we need to use the chain rule.
First, let's find the derivatives of x and y with respect to s and t:
dx/ds = d(s^2 - t^2)/ds
= 2s
dx/dt = d(s^2 - t^2)/dt
= -2t
dy/ds = d(s^2 + t^2)/ds
= 2s
dy/dt = d(s^2 + t^2)/dt
= 2t
Next, let's find the derivatives of z with respect to x and y:
dz/dx = d(sin(2x+y))/dx
= 2cos(2x+y) * dx/dx
= 2cos(2x+y)
dz/dy = d(sin(2x+y))/dy
= cos(2x+y) * dy/dy
= cos(2x+y)
Now, let's find the derivatives of z with respect to s and t:
zₛ = dz/ds
= dz/dx * dx/ds + dz/dy * dy/ds
= 2cos(2x+y) * 2s + cos(2x+y) * 2s
= 4s * cos(2x+y)
zₜ = dz/dt
= dz/dx * dx/dt + dz/dy * dy/dt
= 2cos(2x+y) * (-2t) + cos(2x+y) * 2t
= -4t * cos(2x+y)
(iii) To find the derivatives wₛ and wₜ, we need to use the quotient rule.
First, let's find the derivatives of x, y, and z with respect to s and t:
dx/ds = d(s + t)/ds
= 1
dx/dt = d(s + t)/dt
= 1
dy/ds = d(st)/ds
= t
dy/dt = d(st)/dt
= s
dz/ds = d(s - t)/ds
= 1
dz/dt = d(s - t)/dt
= -1
Now, let's find the derivatives of w with respect to s and t:
wₛ = [dy/ds * (x - z) - y * dz/ds] / (x - z)^2
= [t * (s + t - (s - t)) - st * 1] / (s + t - (s - t))^2
= [2t^2 - st] / (2t)^2
= (2t^2 - st) / 4t^2
= (t(2t - s)) / 4t^2
= (2t - s) / 4t
wₜ = [dy/dt * (x - z) - y * dz/dt] / (x - z)^2
= [s * (s + t - (s - t)) - st * (-1)] / (s + t - (s - t))^2
= [s^2 + st + st] / (2s)^2
= (2st + s^2) / 4s^2
= (s + t) / (2s)
Note: In both cases (i.e., wₛ and wₜ), the expressions can be further simplified, but this is the general form of the derivative.