Calculus-on Manifolds Partial Derivatives and Derivatives, guaranteed and verified 100% Pass

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Partial Derivatives and Derivatives

Let 𝑓: ℝ𝑛 → ℝ and 𝑎 ∈ ℝ𝑛 . We define the 𝒊𝒕𝒉 partial derivative of
𝒇 at 𝒂 as

𝑓(𝑎1 , 𝑎2 , … , 𝑎𝑖 + ℎ, … , 𝑎𝑛 ) − 𝑓 (𝑎1 , 𝑎2 , … , 𝑎𝑛 )
𝐷𝑖 𝑓(𝑎 ) = lim
ℎ→0 ℎ
as long as the limit exists.

Notice that this is just the ordinary derivative of
𝑔(𝑥 ) = 𝑓(𝑎1 , 𝑎2 , … , 𝑥, … , 𝑎𝑛 ) when 𝑥 = 𝑎𝑖 .

Line has slope Line has slope
of 𝐷1 𝑓(𝑎) of 𝐷2 𝑓(𝑎)




So we can calculate a partial derivative by holding all variables “constant”
except the one we are differentiating with respect to.

Ex. Let 𝑓 (𝑥, 𝑦, 𝑧) = sin(𝑥 cos 𝑦) + 𝑥 𝑧 . Find 𝐷1 𝑓 = 𝑓𝑥 , 𝐷2 𝑓 = 𝑓𝑦 , 𝐷3 𝑓 = 𝑓𝑧 .

𝐷1 𝑓 = 𝑓𝑥 = [cos(𝑥 cos 𝑦)](cos 𝑦) + 𝑧𝑥 𝑧−1

𝐷2 𝑓 = 𝑓𝑦 = [cos(𝑥 cos 𝑦)] (−𝑥 sin 𝑦)

𝑧
𝐷3 𝑓 = 𝑓𝑧 = 𝐷3 ((𝑒 ln 𝑥 ) ) = 𝐷3 (𝑒 𝑧 ln 𝑥 )
= (ln 𝑥)𝑒 𝑧 ln 𝑥 = (ln 𝑥)(𝑥 𝑧 ).

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Theorem: If 𝐷𝑖,𝑗 𝑓 and 𝐷𝑗,𝑖 𝑓 are both continuous in an open set
containing 𝑎, then

𝐷𝑖,𝑗 𝑓(𝑎 ) = 𝐷𝑗,𝑖 𝑓 (𝑎 ).

Ex. A function can have a partial derivative everywhere yet not necessarily be
continuous everywhere (in contrast to the statement that if a function has a
derivative at a point, then it is continuous at that point).



Let:
𝑥𝑦 2
𝑓(𝑥, 𝑦) = 𝑥 2+𝑦4 if (𝑥, 𝑦) ≠ (0, 0)
=0 if (𝑥, 𝑦) = (0, 0)

A direct calculation using the quotient rule shows if
(𝑥, 𝑦) ≠ (0, 0), then:
𝑦 2 (𝑦 4 − 𝑥 2 )
𝑓𝑥 =
(𝑥 2 + 𝑦 4 )2
2𝑥𝑦(𝑥 2 − 𝑦 4 )
𝑓𝑦 =
(𝑥 2 + 𝑦 4 )2
If (𝑥, 𝑦) = (0, 0), then:

𝑓(0 + ℎ, 0) − 𝑓 (0, 0) 𝑓(ℎ, 0) − 𝑓(0, 0)
𝑓𝑥 (0, 0) = lim = lim =0
ℎ→0 ℎ ℎ→0 ℎ

𝑓(0, 0 + ℎ) − 𝑓(0, 0) 𝑓 (0, ℎ) − 𝑓(0, 0)
𝑓𝑦 (0, 0) = lim = lim = 0.
ℎ→0 ℎ ℎ→0 ℎ

Thus, 𝑓𝑥 and 𝑓𝑦 exist everywhere.