Chain rule worksheet - solutions
Problem 1
y = 5ecos u ,
u = 3x x3
dy du
dy
= du
.
First, write: dx
dx
dy
Then, notice that while calculating du
you need to use chain rule as well. The outside function is e,
hence:
dy
= 5ecos u (− sin u)
du
In order to calculate du
you need product rule, so:
dx
du
x
3
2 x
= (3 ln 3)x + 3x 3
dx
Together, it gives:
dy
dy du
=
= [5ecos u (− sin u)][(3x ln 3)x3 + 3x2 3x ]
dx
du dx
You can plug u = 3x x3 to the first factor and then you will obtain:
dy
dx
=
dy du
du dx
x x3 )
= [5ecos(3
(− sin(3xx3))][(3x ln 3)x3 + 3x23x]
Problem 2
m = tan
1
u2
+ 7u sin u
5
You need to use chain rule 2 times. You need to apply it first to power function, and then to
tangent. Notice that 7u sin u inside the argument of the function requires product rule.
Remember that while you apply chain rule to the outside function, the argument remains unchanged.
dm
du
= 5 tan
|
4
1
1
2
+ 7u sin u
sec
+ 7u sin u [−2u−3 + (7u cos u + 7 sin u)]
|
{z
}
u2
u2
{z
}|
{z
}
derivative of 1 +7u sin u
derivative of power f unction
derivative of tangent
u2
Problem 3
k=
(2t−1)et
√
,
2 t
t=
1
√
7 2
r
dt
Here you need to use chain rule directly and write dk
= dk
.
dr
dt dr
In order to compute the derivative of k, you need both quotient rule and product rule. First, write:
√
1
dk
[(2t − 1)et ]0 2 t − t− 2 [(2t − 1)et ]
√
=
dt
(2 t)2
Then, since (2t − 1)et consists of two functions, you will need to use product rule, hence the formula
above could be expanded to
√
√
1
1
dk
[(2t − 1)et ]0 2 t − t− 2 [(2t − 1)et ]
[2et + (2t − 1)et ]2 t − t− 2 [(2t − 1)et ]
√
=
=
dt
4t
(2 t)2
Then, notice that
9
1
√
7 2
r
2
= r− 7 . Hence
= − 27 r− 7 .
Together, it gives
dt
dr
√
1
[2et + (2t − 1)et ]2 t − t− 2 [(2t − 1)et ] 2 − 9
dk
=
[− r 7 ].
dr
4t
7
1
You could plug: t = √
7 2 to the first factor and then
r
"
2e
dk
dr
1
√
7
r2
√1 #
√
7
1
1
−1 e r2 2 t− √
+ 2 √
7
7
=
r2
1
−2
r2
" √1 #
7
1
−1 e r2
2 √
7
r2
1
4√
7
r2
9
[− 72 r− 7 ]
Problem 4
(a) f (x) = 5 sin4 3x = 5[sin 3x]4
(b) g(x) = [x3 + 2x ]
5
(c) h(x) = tan(52x )
f 0 (x) = 20[sin 3x]3 (cos 3x)3 = 60[sin 3x]3 (cos 3x)
g 0 (x) = 5[x3 + 2x ]4 (3x2 + 2x ln 2)
h0 (x) = sec2 (52x )[52x (ln 5)2] = 2 sec2 (52x )52x (ln 5)
(d) y = cos(5xex )
dy
dx
(e) m = sin5 (k) = [sin k]5
dm
dk
(f) p =
(r2 +6r )4
r−3
(g) s = esin 2t
(h) f (k) = k sin(2k)
= − sin(5xex )[5ex + 5xex ]
= 5[sin k]4 cos k = 5 sin4 k cos k
4(r2 +6r )3 (2r+6r ln 6)r−3 +3r−4 (r2 +6r )4
(r−3 )2
dp
dr
=
ds
dt
= esin 2t (cos 2t)2 = 2esin 2t cos 2t
f 0 (k) = 1 sin 2k + k(cos 2k)2 = sin 2k + 2k cos 2k
(i) h(x) = sin2 3x cos3 7x = [sin 3x]2 [cos 7x]3
h0 (x) = 2 sin 3x(cos 3x)3[cos 7x]3 + 3[cos 7x]2 (− sin 7x)7[sin 3x]2
(j) s =
h
t−5
et
i3
ds
dt
=3
h
t−5
et
i2
−5t−6 et −et t−5
(et )2