Partial Derivatives & Consequences

Suppose \(f: \Omega \subset \mathbb{R}^n \to \mathbb{R}\). Then \(f\) is differentiable at \(x=a\) if all partial derivatives of \(f\) exist at \(x=a\) and are also continuous at \(a\).

Then \(f\) is diff at \(x=a\) and \(Df(a)(h) = \sum_{j=1}^{n} h_j D_j f(a)\), where \(h = \sum_{j=1}^{n} h_j e_j = (h_1, \dots, h_n)^T\).

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Proof: Let \(a = \sum_{j=1}^{n} a_j e_j\), \(h = \sum_{j=1}^{n} h_j e_j\).

We decompose the difference \(f(a+h) - f(a)\) by adding and subtracting intermediate points:

\begin{align*} f(a+h) - f(a) &= f(a_1+h_1, \dots, a_n+h_n) - f(a_1, \dots, a_n) \\ &= [f(a_1+h_1, \dots, a_n+h_n) - f(a_1+h_1, \dots, a_{n-1}+h_{n-1}, a_n)] \\ &\quad + \dots + [f(a_1+h_1, a_2, \dots, a_n) - f(a_1, \dots, a_n)] \end{align*}

By Mean Value Theorem (MVT) in one variable for each term:

\[ = D_n f(c_n)h_n + D_{n-1} f(c_{n-1})h_{n-1} + \dots + D_1 f(c_1)h_1 \]

where \(c_j\) lies on the line segment. Observe that \(c_j \to a\) as \(h \to 0\) for \(1 \le j \le n\).

By continuity of \(D_j f\) at \(x=a\) (\(1 \le j \le n\)), \(D_j f(c_j) \to D_j f(a)\).

\[ \left| f(a+h) - f(a) - \sum_{j=1}^{n} h_j D_j f(a) \right| = \left| \sum_{j=1}^{n} h_j D_j f(c_j) - \sum_{j=1}^{n} h_j D_j f(a) \right| \] \[ \le \sum_{j=1}^{n} \frac{|h_j|}{|h|} |D_j f(c_j) - D_j f(a)| \le \sum_{j=1}^{n} |D_j f(c_j) - D_j f(a)| \to 0 \text{ as } h \to 0 \]

Since \(\frac{|h_j|}{|h|} \le 1\), the limit goes to 0, establishing differentiability. ∎

Let \(f: \Omega \subset \mathbb{R}^2 \to \mathbb{R}\). If \(f\) is continuously differentiable twice in \(\Omega\), then \(D_1 D_2 f(a,b) = D_2 D_1 f(a,b)\) for all \((a,b) \in \Omega\).

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Proof: Consider a rectangle defined by \((a,b), (a+h,b), (a,b+k), (a+h, b+k)\).

Let \(\Phi(s) = f(s, b+k) - f(s,b)\) for \(s \in [a, a+h]\).
Let \(\Psi(t) = f(a+h, t) - f(a,t)\) for \(t \in [b, b+k]\).

Observe that \(\Phi(a+h) - \Phi(a) = \Psi(b+k) - \Psi(b) = f(a+h, b+k) - f(a+h, b) - f(a, b+k) + f(a, b)\).

\(\Phi\) is continuous in \([a, a+h]\) and differentiable in \((a, a+h)\). \(\Psi\) is continuous in \([b, b+k]\) and differentiable in \((b, b+k)\).

By MVT, \(\exists s_0 \in (a, a+h)\) such that \(\Phi(a+h) - \Phi(a) = h \cdot \Phi'(s_0)\).
And \(\exists t_1 \in (b, b+k)\) such that \(\Psi(b+k) - \Psi(b) = k \cdot \Psi'(t_1)\).

Now, \([D_1 f(s_0, b+k) - D_1 f(s_0, b)] \cdot h = [D_2 f(a+h, t_1) - D_2 f(a, t_1)] \cdot k\).

Applying MVT again to \(D_1 f\) and \(D_2 f\), \(\exists t_0 \in (b, b+k)\) & \(s_1 \in (a, a+h)\) such that:

\[ [D_2 D_1 f(s_0, t_0) \cdot k] \cdot h = [D_1 D_2 f(s_1, t_1) \cdot h] \cdot k \]

For \(h, k \neq 0\), \(D_2 D_1 f(s_0, t_0) = D_1 D_2 f(s_1, t_1)\).

Taking limit as \((h,k) \to (0,0)\), we get \((s_0, t_0) \to (a,b)\) and \((s_1, t_1) \to (a,b)\).
Using the hypothesis that \(f\) is twice continuously differentiable, we conclude \(D_2 D_1 f(a,b) = D_1 D_2 f(a,b)\). ∎

Suppose \(f: \Omega \to \mathbb{R}\) is a differentiable function and \(a,b \in \Omega\) such that the line segment \(a+t(b-a)\), \(0 \le t \le 1\) with the end points \(a\) & \(b\) lies in \(\Omega\).

Then \(\exists c\) on this line segment such that \(f(b) - f(a) = Df(c)(b-a)\).

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Proof: Let \(\phi(t) = f(a+t(b-a))\) for \(t \in [0,1]\).

The map \(t \mapsto a+t(b-a)\) is differentiable on \((0,1)\) & continuous on \([0,1]\). \(f\) is differentiable on \(\Omega\). So \(\phi\) is continuous on \([0,1]\) & differentiable on \((0,1)\).

By 1D MVT, \(\exists t_0 \in (0,1)\) such that \(\phi'(t_0) = \phi(1) - \phi(0) = f(b) - f(a)\).

We have \(\phi'(t) = Df(a+t(b-a)) \cdot D(a+t(b-a))|_{t=t_0} = Df(c) \cdot (b-a)\) where \(c = a+t_0(b-a)\).

So \(f(b) - f(a) = Df(c)(b-a)\). ∎