Calculus on Manifolds Differentiation and Directional Derivatives, guaranteed and verified 100% Pass

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Differentiation and Directional Derivatives

If 𝑓: ℝ → ℝ, we say that 𝑓 is differentiable at 𝑎 ∈ ℝ if
𝑓(𝑎 + ℎ) − 𝑓(𝑎)
lim = 𝑓′(𝑎)
ℎ→0 ℎ

This definition doesn’t make any sense for a function 𝑓: ℝ𝑛 → ℝ𝑚 (in that case,
ℎ ∈ ℝ𝑛 and dividing by a vector is not defined).
However, we can think of any number, 𝑓 ′ (𝑎 ), as defining a linear transformation
𝜆 of ℝ into ℝ by:
𝜆: ℝ → ℝ
𝜆(ℎ) = (𝑓 ′ (𝑎 ))ℎ.

So we could reformulate our definition of the derivative, 𝑓 ′ (𝑎 ), by saying:
𝑓(𝑎 + ℎ) − 𝑓(𝑎 ) − 𝑓 ′ (𝑎)ℎ
lim =0
ℎ→0 ℎ
Or
𝑓(𝑎 + ℎ) − 𝑓(𝑎 ) − 𝜆(ℎ)
lim =0
ℎ→0 ℎ


Thus, we say a function 𝑓: ℝ → ℝ is differentiable at 𝑎 ∈ ℝ if there is a linear
transformation 𝜆: ℝ → ℝ such that:

𝑓(𝑎 + ℎ) − 𝑓(𝑎 ) − 𝜆(ℎ)
lim =0
ℎ→0 ℎ

Note: Any linear transformation, 𝜆, of ℝ into ℝ, 𝜆: ℝ → ℝ, is just
multiplication by a fixed number; 𝜆(ℎ) = 𝑝ℎ; 𝑝 ∈ ℝ.

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Now we can generalize this definition to: 𝑓: ℝ𝑛 → ℝ𝑚 .

Def. A function 𝑓: ℝ𝑛 → ℝ𝑚 is differentiable at 𝑎 ∈ ℝ𝑛 if there is a linear
transformation, 𝜆: ℝ𝑛 → ℝ𝑚 , such that:
|𝑓(𝑎 + ℎ) − 𝑓(𝑎 ) − 𝜆(ℎ)|
lim = 0.
ℎ→0 |ℎ|

Notice that (𝑓(𝑎 + ℎ) − 𝑓(𝑎 ) − 𝜆(ℎ)) ∈ ℝ𝑚 and ℎ ∈ ℝ𝑛 .
If this limit is 0, then we say: 𝐷𝑓 (𝑎 ) = 𝜆.



Theorem: If 𝑓: ℝ𝑛 → ℝ𝑚 is differentiable at 𝑎 ∈ ℝ𝑛 , then there is a
unique linear transformation, 𝜆: ℝ𝑛 → ℝ𝑚 , such that:

|𝑓(𝑎 + ℎ) − 𝑓(𝑎 ) − 𝜆(ℎ)|
lim = 0.
ℎ→0 |ℎ|

Proof: Suppose 𝜏: ℝ𝑛 → ℝ𝑚 is a linear transformation that also satisfies

|𝑓(𝑎 + ℎ) − 𝑓(𝑎 ) − 𝜏(ℎ)|
lim = 0.
ℎ→0 |ℎ|

Then we have:
|𝜆(ℎ) − 𝜏(ℎ)|
0 ≤ lim
ℎ→0 |ℎ|

|(𝜆(ℎ) − 𝑓(𝑎 + ℎ) + 𝑓 (𝑎 )) + (𝑓(𝑎 + ℎ) − 𝑓 (𝑎 ) − 𝜏(ℎ))|
= lim
ℎ→0 |ℎ|

|𝑓(𝑎 + ℎ) − 𝑓(𝑎 ) − 𝜆(ℎ)| |𝑓 (𝑎 + ℎ) − 𝑓(𝑎 ) − 𝜏(ℎ)|
≤ lim + lim
ℎ→0 |ℎ| ℎ→0 |ℎ|
= 0 + 0 = 0. ⇒ 𝜆(ℎ) = 𝜏(ℎ).

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Ex. Let 𝑓: ℝ2 → ℝ2 by 𝑓(𝑥, 𝑦) = (𝑥𝑦, 𝑥 + 2𝑦). Using the definition of the
derivative, show that:
0 0
𝐷𝑓 (0, 0) = ( ).
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|𝑓(⃗0+ℎ)−𝑓(⃗0)−𝜆(ℎ)| 0 0
We must show that lim = 0, where 𝜆 = ( ).
ℎ→0 |ℎ| 1 2
If we let ℎ = (ℎ1 , ℎ2 ) then:

0 0 ℎ1
⃗ ⃗
|𝑓(0 + ℎ) − 𝑓(0) − 𝜆(ℎ)| |(ℎ1 ℎ2 , ℎ1 + 2ℎ2 ) − ( ) ( )|
1 2 ℎ2
lim = lim
ℎ→0 |ℎ| ℎ→0 |ℎ |

|(ℎ1 ℎ2 , ℎ1 + 2ℎ2 ) − (0, ℎ1 + 2ℎ2 )| |ℎ1 ℎ2 |
= lim = lim
ℎ→0 |ℎ| ℎ→0 √ℎ2 + ℎ2
1 2

2 2
Notice that (ℎ1 + ℎ2 )2 = ℎ1 + 2ℎ1 ℎ2 + ℎ2 ≥ 0

ℎ12 + ℎ22 ≥ −2ℎ1 ℎ2
ℎ12 +ℎ22
≥ |ℎ1 ℎ2 | .
2

So:
ℎ2 2
1 +ℎ2
|ℎ1ℎ2 | 1
0 ≤ lim ≤ lim 2
= lim 2 √ℎ12 + ℎ22 = 0
ℎ→0 √ 2 ℎ→0 √ 2 2 ℎ→0
ℎ1+ℎ22 ℎ1+ℎ2

⃗ +ℎ)−𝑓(0
|𝑓(0 ⃗ )−𝜆(ℎ)|
Thus lim = 0 and:
ℎ→0 |ℎ|


0 0
𝐷𝑓 (0, 0) = ( ).
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