Let \(f: \Omega \subset \mathbb{R}^n \to \mathbb{R}\), where \(a \in \Omega\) (open).
\(f\) is said to be differentiable in the \(j\)-th variable at \(x=a\) if
\[ \lim_{t \to 0} \frac{f(a+te_j) - f(a)}{t} \text{ exists.} \](where \((a_1, \dots, a_j+t, \dots, a_n) = a + te_j\)).
It is called the partial derivative of \(f\) at \(x=a\), denoted as \(D_j f(a)\):
\[ D_j f(a) := \lim_{t \to 0} \frac{f(a+te_j) - f(a)}{t} \]If \(g(t) = f(a+te_j)\) (\(t \in \mathbb{R}\)), then \(g(0) = f(a)\). And \(\frac{g(t)-g(0)}{t-0} = \frac{f(a+te_j)-f(a)}{t}\).
Suppose \(f\) admits a maximum or minimum at an interior point \(a\) of \(A\), where \(f: A \to \mathbb{R}\). If the \(j\)-th partial derivative of \(f\) exists at \(x=a\), then it must be zero.
If \(f\) is differentiable at \(x=a\), then for \(1 \le j \le n\), \(D_j f(a)\) exists.
Proof: We have \(\lim_{h \to 0} \frac{|f(a+h) - f(a) - Df(a)(h)|}{|h|} = 0\).
Substitute \(h = te_j\). As \(t \to 0\), \(h \to 0\).
\[ \lim_{t \to 0} \frac{|f(a+te_j) - f(a) - Df(a)(te_j)|}{|t|} = 0 \] \[ \implies \lim_{t \to 0} \left| \frac{f(a+te_j) - f(a) - tDf(a)(e_j)}{t} \right| = 0 \] \[ \implies \lim_{t \to 0} \left| \frac{f(a+te_j) - f(a)}{t} - Df(a)(e_j) \right| = 0 \]Thus, \(\lim_{t \to 0} \frac{f(a+te_j) - f(a)}{t} = Df(a)(e_j)\). Hence, \(D_j f(a) = Df(a)(e_j)\) (Partial derivative w.r.t \(j\)-th coord). ∎
\(f'(a) = [Df(a)(e_1) \dots Df(a)(e_n)]\).
The converse is not true in general. The existence of partial derivatives does not guarantee total differentiability.
English explanation and missing steps
\(D_jf(a)\) changes only coordinate \(j\): it is the derivative of the one-variable slice \(t\mapsto f(a+te_j)\) at zero. If the full derivative exists, \(D_jf(a)=Df(a)e_j\). The converse fails: separate axis limits do not control paths between axes.
For vector output, each \(D_jf(a)\) is a vector and becomes column \(j\) of the Jacobian. To test full differentiability at a suspect point, build the candidate matrix from partials and evaluate the remainder divided by \(|h|\) along general paths.
Hinglish mein poora explanation
Partial derivative \(D_jf(a)\) nikalte waqt sirf coordinate \(j\) badalta hai; baaki fixed rahte hain. Agar full derivative hai, to \(D_jf(a)=Df(a)e_j\). Lekin sab partials milne se full differentiability automatically nahi milti, kyunki beech ke paths check nahi hue.
Vector output mein \(D_jf(a)\) ek vector hai, Jacobian ka column \(j\). Doubtful origin problems mein partials se candidate matrix banao, phir general \(h\) par remainder ko \(|h|\) se divide karke limit test karo.
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 \(h = \sum_{j=1}^{n} h_j e_j = (h_1, \dots, h_n)^T\).
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. ∎
English explanation and missing steps
The useful precise hypothesis is that each partial derivative exists in a neighborhood of \(a\) and is continuous at \(a\). Move from \(a\) to \(a+h\) one coordinate at a time. The one-variable mean value theorem turns each increment into \(h_jD_jf(\xi_j)\). As \(h\to0\), every intermediate \(\xi_j\to a\), so continuity replaces \(D_jf(\xi_j)\) by \(D_jf(a)+o(1)\). Summing finitely many terms yields a remainder \(o(|h|)\).
Continuity of partials is sufficient, not necessary; differentiable functions can have discontinuous partial derivatives.
Hinglish mein poora explanation
Precise condition: saare partial derivatives \(a\) ke neighborhood mein exist karen aur \(a\) par continuous hon. \(a\) se \(a+h\) tak ek-ek coordinate change karo. Har step par one-variable MVT se difference \(h_jD_jf(\xi_j)\) banta hai. \(h\to0\) par \(\xi_j\to a\), to continuity se \(D_jf(\xi_j)=D_jf(a)+o(1)\). Finite sum ka error \(o(|h|)\) hai.
Yaad rakho: continuous partials sufficient hain, necessary nahi. Kuch differentiable functions ke partials origin par discontinuous hote hain.
\(f: \Omega \subset \mathbb{R}^n \to \mathbb{R}^m\) is continuously differentiable (or \(C^1\)) at \(a \iff f\) is diff at \(a\) and \(f'(a) = (D_1 f(a), \dots, D_n f(a))\) is continuous.
We will also denote \(D_i f(a) = \frac{\partial f}{\partial x_i}(a)\).
\(f: \mathbb{R}^n \to \mathbb{R}\) is said to be \(k\)-times continuously differentiable (\(C^k\)) if all partial derivatives of \(f\) up to (including) order \(k\) exist and are continuous in \(\Omega\).
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 \((a,b) \in \Omega\).
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 \(D_2 D_1 f(a,b) = D_1 D_2 f(a,b)\). ∎
English explanation and missing steps
\(C^1\) means all first partials exist on an open set and vary continuously there. \(C^k\) requires continuous partials through order \(k\). For mixed partials, Clairaut's theorem needs suitable continuity of the second partials near the point; mere existence of \(D_{12}\) and \(D_{21}\) does not guarantee equality.
The order notation matters: \(D_1D_2f\) means apply \(D_2\) first, then \(D_1\). In examples, calculate each first partial on the axes before differentiating it again at the origin.
Hinglish mein poora explanation
\(C^1\) ka matlab open set mein first partials exist aur continuous hain. \(C^k\) mein order \(k\) tak continuous partials chahiye. Mixed partials \(D_{12}\) aur \(D_{21}\) ko equal bolne ke liye theorem ki continuity condition check karo; sirf dono ka exist karna enough nahi.
Notation mein \(D_1D_2f\) par right wala \(D_2\) pehle lagta hai. Origin wale examples mein pehle axis par first partial ka formula nikalo, phir uska origin par derivative lo.
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 \(f(b) - f(a) = Df(c)(b-a)\).
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 \(f(b) - f(a) = Df(c)(b-a)\). ∎
English explanation and missing steps
For scalar \(f\), restrict to the line segment: \(\phi(t)=f(a+t(b-a))\). Chain rule gives \(\phi'(t)=Df(a+t(b-a))(b-a)\). The ordinary one-variable MVT gives \(\phi(1)-\phi(0)=\phi'(t_*)\), which is the stated formula.
Do not use one scalar \(c\) automatically for a vector-valued map; different components can require different intermediate points. The reliable vector estimate is \(|f(b)-f(a)|\le\sup_{segment}\|Df\|\,|b-a|\) under suitable regularity.
Hinglish mein poora explanation
Line segment ko one-variable path banao: \(\phi(t)=f(a+t(b-a))\). Chain rule se \(\phi'(t)=Df(a+t(b-a))(b-a)\). Ab usual one-variable Mean Value Theorem lagao; isi se lecture ka formula milta hai.
Vector-valued map ke liye ek hi intermediate point \(c\) sab components ke liye zaroori nahi hota. Wahan derivative norm ka bound use karke \(|f(b)-f(a)|\) estimate karna safe method hai.
\(D_j f(a) = Df(a)(e_j)\), when derivative of \(f\) exists at \(x=a\).
\(\phi(t) = f(a+te_j)\) is diff at \(t=0\), when \(D_j f(a)\) exists.
\(\frac{d}{dt}|_{t=0} \phi(t) = Df(a)(e_j)\).
Let \(f: \Omega \to \mathbb{R}\), \(a \in \Omega\), \(0 \neq v \in \mathbb{R}^n\). Define the directional derivative of \(f\) at \(a\) w.r.t \(v\) as:
\[ D_v f(a) := \lim_{t \to 0} \frac{f(a+tv) - f(a)}{t} = \left. \frac{d}{dt} \right|_{t=0} f(a+tv) \]if it exists. If \(Df(a)\) exists, \(D_v f(a)\) exists and equals \(Df(a)(v)\).
- Suppose \(f: \Omega \to \mathbb{R}\) is diff at \(x=a\). Then \(D_v f(a)\) exists & is equal to \(Df(a)(v)\).
- Suppose \(f\) is Ctsly diff and say \(\exists\) diff fun \(g: \mathbb{R} \to \mathbb{R}\) st \(g(f(x)) = x \forall x \in \Omega\).
English explanation and missing steps
Directional derivative is the derivative of \(t\mapsto f(a+tv)\) at \(t=0\). If \(f\) is differentiable, the chain rule gives \(D_vf(a)=Df(a)v\), hence \(D_{u+v}f(a)=D_uf(a)+D_vf(a)\) and \(D_{cv}f(a)=cD_vf(a)\).
Without full differentiability, directional derivatives may all exist but fail additivity. For unit direction \(v\), \(D_vf=\nabla f\cdot v\) when the gradient exists through full differentiability. For nonunit \(v\), the derivative also includes the speed \(|v|\).
Hinglish mein poora explanation
Direction \(v\) mein derivative, one-variable function \(t\mapsto f(a+tv)\) ka derivative at \(t=0\) hai. Full differentiability ho to chain rule se \(D_vf(a)=Df(a)v\), isliye directions mein additivity aur scalar rule milte hain.
Sirf directional derivatives exist karne se additivity guaranteed nahi. Unit vector ke liye gradient ka dot product rate of change deta hai. Nonunit \(v\) mein \(|v|\) wali speed bhi included hoti hai.