Partial Derivatives Worksheet
1. Find all the first and second partial derivatives of f (x, y) = sin x cos y
Solution:
fx = cos x cos y
fy = − sin x sin y
fxx = − sin x cos y
fxy = − cos x sin y
fyy = − sin x cos y
2. Find all the first and second partial derivatives of f (x, y) = x3 y +
x2 y 2 + y 5 + 2 − 23xy
Solution:
fx = 3x2 y + 2xy 2 − 23y
fy = x3 + 2x2 y + 5y 4 − 23x
fxx = 6xy + 2y 2
fxy = 3x3 + 4xy − 23
fyy = 2x2 + 20y 3
3. Find all the first and second partial derivatives of f (x, y) = x3 sin(x +
y)ey
Solution:
fx = 3x2 sin(x + y)ey + x3 cos(x + y)ey
fy = x3 cos(x + y)ey + x3 sin(x + y)ey
fxx = 6x sin(x + y)ey + 6x2 cos(x + y)ey − x3 sin(x + y)ey
fxy = 3x2 cos(x + y)ey + 3x2 sin(x + y)ey − x3 sin(x + y)ey + x3 cos(x + y)ey
fyy = 2x3 cos(x + y)ey
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4. Find all the first and second partial derivatives of f (x, y) = x2 + y 2
Solution:
fx = 2x
fy = 2y
fxx = 2
fxy = 0
fyy = 2
5. Find all the first and second partial derivatives of f (x, y) =
p
1 − x2 − y 2
Solution:
−x
fx = p
1 − x2 − y 2
−y
fy = p
1 − x2 − y 2
−1
x2
fxx = p
−
1 − x2 − y 2 (1 − x2 − y 2 )3/2
y2 − 1
=
(1 − x2 − y 2 )3/2
−xy
fxy =
(1 − x2 − y 2 )3/2
−1
y2
fyy = p
−
1 − x2 − y 2 (1 − x2 − y 2 )3/2
x2 − 1
=
(1 − x2 − y 2 )3/2
6. Find all the first and second partial derivatives of f (x, y) =
Solution:
2
1
x+y 2 +1
fx =
fy =
fxx =
fxy =
fyy =
=
−1
(x + y 2 + 1)2
−2y
(x + y 2 + 1)2
2
(x + y 2 + 1)3
4y
(x + y 2 + 1)3
8y 2
−2
+
(x + y 2 + 1)2 (x + y 2 + 1)3
−2x + 6y 2 − 2
(x + y 2 + 1)3
7. Find an equation for the plane tangent to the surface f (x, y) = xy 2 at
(2, 3).
Solution:
fx = y 2
fy = 2xy
fx (2, 3) = 9
fy (2, 3) = 12
This gives us the plane determined by the equation
9(x − 2) + 12(y − 3) = z − 18
8. Find an equation for the plane tangent to the surface f (x, y) = sin x cos y
at π2 , π2 .
Solution:
fx = cos x cos y
fy = − sin x sin y
π π fx
,
=0
2 2
π
π
fy
,
= −1
2 2
3
This gives us the plane determined by the equation
0(x − π/2) + 1(y − π/2) = z − 0
9. Find an equation for the plane tangent to the surface f (x, y) = exy at
(1, 2).
Solution:
fx = yexy
fy = xexy
fx (1, 2) = 2e2
fy (1, 2) = e2
This gives us the plane determined by the equation
2e2 (x − 1) + e2 (y − 2) = z − e2
10. Find an equation for the plane tangent to the surface f (x, y) = xey +
x sin(x + y) at (0, 0).
Solution:
fx = ey + sin(x + y) + x cos(x + y)
fy = xey + x cos(x + y)
fx (0, 0) = 1
fy (0, 0) = 0
This gives us the plane determined by the equation
1(x − 0) + 0(y − 0) = z − 0
11. Given that f (r, θ) = r3 cos θ, find
∂f
∂x
and
∂f
∂y .
Solution:
There are a few ways
p of doing this. The first yis to replace all the
instances of r with x2 + y 2 and θ with arctan x . Alternatively, we
can notice that
r3 cos θ = r2 · r cos θ
= (x2 + y 2 )x
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This makes the derivatives really easy to take. (The other option is to
use the chain rule, which takes way longer.) Using the first technique,
we get fx = 3x2 + y 2 and fy = 2xy.
12. Let g(u, v) = f (x(u, v), y(u, v)), where x(u, v) = u2 + v 2 , y(u, v) =
u2 − v 2 . Then, given the following values for the partial derivatives of
f , calculate gu (1, 1) and gv (2, 1)
x
0
2
0
1
5
3
2
y fx (x, y) fy (x, y)
0
1
2
0
3
4
2
5
6
1
9
10
3
−1
−2
5
−3
−4
1
−5
−6
Solution:
First off, we note that x(1, 1) = 2, y(1, 1) = 0 and x(2, 1) = 5, y(2, 1) =
3
gu (1, 1) = fx (2, 0)xu (1, 1) + fy (2, 0)yu (1, 1)
=3·2+4·2
= 14
gv (2, 1) = fx (5, 3)xv (2, 1) + fy (5, 3)yv (2, 1)
= −1 · 2 + (−2) · (−2)
=2
13. Consider the surface defined by the equation xy − yz + exz = 3. Is
there a function f (x, y) such that in a neighborhood of (0, −1, −3), the
surface can be described by z = f (x, y)? What are fx (0, −1, −3) and
fy (0, −1, −3)? Find the equation for the plane tangent to the surface
at this point.
Solution:
In order to apply the implicit function theorem here, we must verify
∂
that ∂z
g, where g is the equation defining the surface, is non-zero.
∂
(xy − yz + exz ) = −y + xexz
∂z
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At our point, this has value 1. Thus we can apply the implicit function
theorem.
gx
gz
= −4
gy
fy =
gz
=3
fx =
This gives us the equation for the plane z + 3 = −4x + 3(y + 1).
∂f
2
2
14. If ∂f
∂x = Ax + 2xy + y, ∂y = Bx + Cy + x, then what values can
A, B, C take?
Solution:
Applying Clairaut’s equation, we must have
∂
∂
Ax2 + 2xy + y =
Bx2 + Cy + x
∂y
∂x
2x + 1 = 2Bx + 1
B=1
Beyond this, however, there are no requirements for A and C.
15. Let g(x, y) be a function and define f (u, v) = g(uv, u − v). Express
the following in terms of u, v, gx , gy , gxx , gxy , gyy :
∂f ∂f ∂ 2 f
∂u ∂v ∂u∂v
Solution:
∂f
∂
= gx (uv, u − v) (uv)+
∂u
∂u
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