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October 26, 2019 MAT186 – Week 6 Justin Ko
1 Applications of the Chain Rule
We go over several examples of applications of the chain rule to compute derivatives of more compli-
cated functions.
Chain Rule: If z = f(y) and y = g(x) then
d f(g(x)) = f′(g(x))·g′(x) or equivalently dz = dz · dy.
dx dx dy dx
The chain rule is used as the main tool to solve the following classes for problems:
1. Derivatives of Inverse Functions: The chain rule is used to derive the derivative of the inverse
function formula. If y = f−1(x) and x = f(y) then
d f−1(x) = 1 or equivalently dy = 1 .
dx f′(f−1(x)) dx dx
dy
2. Implicit Differentiation: The chain rule can be used to compute derivatives of implicit functions
F(x,y(x)) = 0
where F is a function of two variables x and y.
3. Logarithmic Differentiation: By first taking the logarithm of both sides, we can compute deriva-
tives of
g(x)
y(x) = f(x) .
4. Related Rates: There are word problems where both y and x depend on some related variable t.
The goal is to compute the rate of change of y(x) with respect to t.
1.1 Example Problems
1.1.1 Derivatives of Inverse Functions
Problem 1.1. (⋆⋆) Prove the formula for the derivative of the inverse function
d f−1(x) = 1 .
dx f′(f−1(x))
Solution 1.1. By the cancellation laws, we have
(f ◦ f−1)(x) = x.
Differentiating both sides and using the chain rule, we have
d f(f−1(x)) = d x ⇒ f′(f−1(x))· d f−1(x) = 1 ⇒ d f−1(x) = 1 .
dx dx dx dx f′(f−1(x))
Remark. Alternatively, if we let y(x) = f−1(x) and x(y) = f(y), then x(y(x)) = f(f−1(x)) = x, so
differentiating both sides with respect to x implies
1 = dx · dy =⇒ dy = 1 .
dy dx dx dx
dy
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October 26, 2019 MAT186 – Week 6 Justin Ko
Problem 1.2. (⋆) Let f(x) = √x. Find d f−1(x).
dx
′ 1 −1/2 1 −1 2
Solution 1.2. Since f (x) = 2x = 2√x and f (x) = x , the formula for the derivative of the
inverse implies that p
d f−1(x) = 1 =2 f−1(x)=2x.
dx f′(f−1(x))
−1 √
Remark. Using Leibniz notation, if we set y = f (x), then x = f(y) = y. Therefore,
dy 1 1 √
= dx = 1 =2 y=2x,
dx √
dy 2 y
√ 2
since x = y =⇒ y =x . This method is exactly the same as the one above. We just used different
notation to convey the same idea.
Problem 1.3. (⋆⋆) Let f(x) = x+sin(x). Find (f−1)′(0).
Solution 1.3. Notice that x + sin(x) is one-to-one on R, but its inverse is impossible to express in
terms of functions we have encountered so far. However, we can still find the derivative of the inverse
using the formula for the derivative of the inverse function.
Notice that f(0) = 0 + sin(0) = 0, so we have f−1(0) = 0. Since f′(x) = 1 + cos(x), the formula
for the inverse derivative implies
(f−1)′(0) = 1 = 1 = 1 = 1.
f′(f−1(0)) f′(0) 1+cos(0) 2
x
Problem 1.4. (⋆⋆) Consider the function f(x) = xe restricted to the domain x ≥ 0. The Lambert
W function is defined as the inverse function of f,
W(x)=f−1(x).
Find W′(x).
′ x x
Solution 1.4. Since f (x) = e +xe , the formula for the derivative of inverse function implies
W′(x) = 1 = 1 = 1 ,
f′(W(x)) W(x) W(x) W(x)
e +W(x)e e +x
W(x)
since W(x)e =f(W(x))=x.
Problem 1.5. (⋆⋆) Using the formula for the derivative of the inverse function, show that
d sin−1(x) = √ 1 .
dx 2
1−x
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October 26, 2019 MAT186 – Week 6 Justin Ko
Solution 1.5. We will use the formula for the derivative of inverse functions
d f−1(x) = 1 = 1
dx f′(f−1(x)) ( d f ◦ f−1)(x)
dx
−1 −1 d
with f(x) = sin(x). Since f (x) = sin (x) and dx sin(x) = cos(x), the formula implies
d sin−1(x) = 1 .
dx cos(sin−1(x))
−1
Wenowwantto simplify the function cos(sin (x)) without using trigonometric identities. This type
of problem was introduced in Week 1:
Geometric Solution:
Case x ≥ 0: We first consider the case such that x > 0 on our domain. On this region, we have
−1 π
θ = sin (x) ∈ [0, 2] (the first quadrant). The triangle corresponding to sin(θ) = x in the first
quadrant is given by
1 x
θ √
1−x2
From this triangle, we see
1 = 1 =√ 1 for x ∈ [0,1).
cos(sin−1(x)) cos(θ) 1−x2
Case x < 0: We first consider the case such that x > 0 on our domain. On this region, we have
θ = sin−1(x) ∈ [−π,0] (the fourth quadrant). The triangle corresponding to sin(θ) = x in the fourth
2
quadrant is given by
√1−x2
θ
x
1
Notice that x < 0, so the side opposite the θ is positive. From this triangle, we see
1 = 1 =√ 1 for x ∈ (−1,0).
−1 cos(θ) 2
cos(sin (x)) 1−x
Combining the two domains, we have
d −1 √ 1
dx sin (x) = 2 for x ∈ (−1,1).
1−x
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October 26, 2019 MAT186 – Week 6 Justin Ko
−1
Remark. Using Leibniz notation, if we set y = sin (x), then x = sin(y). Therefore,
dy = 1 = 1 = 1 .
dx dx cos(y) cos(sin−1(x))
dy
−1
Wenowsimplify cos(sin (x)) without using trigonometric functions like above.
Problem 1.6. (⋆⋆) Using the formula for the derivative of the inverse function, show that
d sec−1(x) = √1 .
dx |x| x2 −1
Solution 1.6. We will use the formula
d f−1(x) = 1 = 1
dx f′(f−1(x)) ( d f ◦ f−1)(x)
dx
−1 −1 d
with f(x) = sec(x). Since f (x) = sec (x) and dx sec(x) = sec(x)tan(x), the formula implies
d sec−1(x) = 1 .
−1 −1
dx sec(sec (x))tan(sec (x))
−1 −1
Wenowwantto simplify the function sec(sec (x))tan(sec (x)) without using trigonometric identi-
ties. This type of problem was introduced in Week 1:
Geometric Solution:
Case x > 0: We first consider the case such that x > 0 on our domain. On this region, we have
−1 π
θ = sec (x) ∈ [0, 2] (the first quadrant). The triangle corresponding to sec(θ) = x in the first
quadrant is given by
x √ 2
x −1
θ 1
From this triangle, we see
1 = 1 = √1 for x ∈ (1,∞).
−1 −1 2
sec(sec (x))tan(sec (x)) sec(θ)tan(θ) x x −1
Case x < 0: We now consider the case such that x < 0 on our domain. On this region, we have
−1 π
θ = sec (x) ∈ [2,π] (the second quadrant). The triangle corresponding to sec(θ) = x in the second
quadrant is given by
√ 2 −x
x −1
−1 θ
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