Intuition The big pictureIn single-variable calculus, a function is differentiable at a a a line there. In multiple variables, differentiability means the surface has a tangent plane that genuinely hugs the graph — the function looks linear when you zoom in.
Here is the trap: having partial derivatives is NOT enough. You can have ∂ f / ∂ x \partial f/\partial x ∂ f / ∂ x ∂ f / ∂ y \partial f/\partial y ∂ f / ∂ y stronger condition than "partials exist".
There is a strict hierarchy. Each implies the one below, never the reverse.
continuously diff. ( C 1 ) ⇒ differentiable ⇒ partials exist \text{continuously diff. } (C^1) \;\Rightarrow\; \text{differentiable} \;\Rightarrow\; \text{partials exist} continuously diff. ( C 1 ) ⇒ differentiable ⇒ partials exist differentiable ⇒ continuous \text{differentiable} \;\Rightarrow\; \text{continuous} differentiable ⇒ continuous
Definition Partial derivativeFor f : R 2 → R f:\mathbb{R}^2\to\mathbb{R} f : R 2 → R f x ( a , b ) = lim h → 0 f ( a + h , b ) − f ( a , b ) h , f y ( a , b ) = lim h → 0 f ( a , b + h ) − f ( a , b ) h . f_x(a,b)=\lim_{h\to 0}\frac{f(a+h,b)-f(a,b)}{h},\qquad f_y(a,b)=\lim_{h\to 0}\frac{f(a,b+h)-f(a,b)}{h}. f x ( a , b ) = lim h → 0 h f ( a + h , b ) − f ( a , b ) , f y ( a , b ) = lim h → 0 h f ( a , b + h ) − f ( a , b ) . one straight direction (along the axis). It is blind to everything happening off that line.
Definition Differentiability (the real definition)f f f differentiable at a = ( a , b ) \mathbf{a}=(a,b) a = ( a , b ) ∇ f ( a ) \nabla f(\mathbf a) ∇ f ( a ) lim h → 0 ∣ f ( a + h ) − f ( a ) − ∇ f ( a ) ⋅ h ∣ ∥ h ∥ = 0. \lim_{\mathbf{h}\to\mathbf 0}\frac{\big|\,f(\mathbf a+\mathbf h)-f(\mathbf a)-\nabla f(\mathbf a)\cdot\mathbf h\,\big|}{\|\mathbf h\|}=0. lim h → 0 ∥ h ∥ f ( a + h ) − f ( a ) − ∇ f ( a ) ⋅ h = 0. error of the linear approximation must shrink faster than the distance ∥ h ∥ \|\mathbf h\| ∥ h ∥ every direction at once.
We want a tangent plane L ( x ) = f ( a ) + m ⋅ ( x − a ) L(\mathbf x)=f(\mathbf a)+\mathbf m\cdot(\mathbf x-\mathbf a) L ( x ) = f ( a ) + m ⋅ ( x − a ) f f f E ( h ) = f ( a + h ) − [ f ( a ) + m ⋅ h ] ⏟ plane . E(\mathbf h)=f(\mathbf a+\mathbf h)-\underbrace{\big[f(\mathbf a)+\mathbf m\cdot\mathbf h\big]}_{\text{plane}}. E ( h ) = f ( a + h ) − plane [ f ( a ) + m ⋅ h ] .
For a tangent plane we don't just want E → 0 E\to 0 E → 0 E E E negligible compared to the step size ∥ h ∥ \|\mathbf h\| ∥ h ∥ higher-order error:
E ( h ) = o ( ∥ h ∥ ) ⟺ E ( h ) ∥ h ∥ → 0. E(\mathbf h)=o(\|\mathbf h\|)\quad\Longleftrightarrow\quad \frac{E(\mathbf h)}{\|\mathbf h\|}\to 0. E ( h ) = o ( ∥ h ∥ ) ⟺ ∥ h ∥ E ( h ) → 0.
∥ h ∥ \|\mathbf h\| ∥ h ∥ E → 0 E\to 0 E → 0 ∥ h ∥ \|\mathbf h\| ∥ h ∥ quadratically-fast . That's exactly what makes it a tangent plane and not just any plane that touches the point.
Which m \mathbf m m Approach along the x x x h = ( h , 0 ) \mathbf h=(h,0) h = ( h , 0 ) ∥ h ∥ = ∣ h ∣ \|\mathbf h\|=|h| ∥ h ∥ = ∣ h ∣ f ( a + h , b ) − f ( a , b ) − m 1 h ∣ h ∣ → 0 ⇒ m 1 = f x ( a , b ) . \frac{f(a+h,b)-f(a,b)-m_1 h}{|h|}\to 0 \;\Rightarrow\; m_1=f_x(a,b). ∣ h ∣ f ( a + h , b ) − f ( a , b ) − m 1 h → 0 ⇒ m 1 = f x ( a , b ) . m 2 = f y ( a , b ) m_2=f_y(a,b) m 2 = f y ( a , b ) if f f f ∇ f = ( f x , f y ) \nabla f=(f_x,f_y) ∇ f = ( f x , f y )
f ( a + h ) = f ( a ) + ∇ f ( a ) ⋅ h + o ( ∥ h ∥ ) \boxed{\,f(\mathbf a+\mathbf h)=f(\mathbf a)+\nabla f(\mathbf a)\cdot\mathbf h+o(\|\mathbf h\|)\,} f ( a + h ) = f ( a ) + ∇ f ( a ) ⋅ h + o ( ∥ h ∥ )
Suppose f f f a \mathbf a a f ( a + h ) − f ( a ) = ∇ f ( a ) ⋅ h + E ( h ) , E = o ( ∥ h ∥ ) . f(\mathbf a+\mathbf h)-f(\mathbf a)=\nabla f(\mathbf a)\cdot\mathbf h+E(\mathbf h),\quad E=o(\|\mathbf h\|). f ( a + h ) − f ( a ) = ∇ f ( a ) ⋅ h + E ( h ) , E = o ( ∥ h ∥ ) . h → 0 \mathbf h\to\mathbf 0 h → 0 → 0 \to 0 → 0 h \mathbf h h E → 0 E\to 0 E → 0 E / ∥ h ∥ → 0 E/\|\mathbf h\|\to0 E /∥ h ∥ → 0 E → 0 E\to0 E → 0 f ( a + h ) → f ( a ) f(\mathbf a+\mathbf h)\to f(\mathbf a) f ( a + h ) → f ( a ) Continuous. ■ \blacksquare ■
Worked example Example 1 — Confirm differentiability the easy wayf ( x , y ) = x 2 y + sin y f(x,y)=x^2 y + \sin y f ( x , y ) = x 2 y + sin y ( 1 , 0 ) (1,0) ( 1 , 0 )
f x = 2 x y f_x=2xy f x = 2 x y f y = x 2 + cos y f_y=x^2+\cos y f y = x 2 + cos y Why this step? Compute partials treating the other variable as constant.Both are continuous everywhere (products/sums of continuous functions). Why? C 1 ⇒ C^1\Rightarrow C 1 ⇒
∇ f ( 1 , 0 ) = ( 0 , 1 + 1 ) = ( 0 , 2 ) \nabla f(1,0)=(0,\;1+1)=(0,2) ∇ f ( 1 , 0 ) = ( 0 , 1 + 1 ) = ( 0 , 2 ) z = 0 + 0 ( x − 1 ) + 2 ( y − 0 ) = 2 y z=0+0(x-1)+2(y-0)=2y z = 0 + 0 ( x − 1 ) + 2 ( y − 0 ) = 2 y
Worked example Example 2 — Partials exist but NOT differentiablef ( x , y ) = { x y x 2 + y 2 ( x , y ) ≠ ( 0 , 0 ) 0 ( 0 , 0 ) f(x,y)=\begin{cases}\dfrac{xy}{x^2+y^2}&(x,y)\neq(0,0)\\[2mm]0&(0,0)\end{cases} f ( x , y ) = ⎩ ⎨ ⎧ x 2 + y 2 x y 0 ( x , y ) = ( 0 , 0 ) ( 0 , 0 )
Partials at origin: along y = 0 y=0 y = 0 f ( x , 0 ) = 0 f(x,0)=0 f ( x , 0 ) = 0 f x ( 0 , 0 ) = 0 f_x(0,0)=0 f x ( 0 , 0 ) = 0 f y ( 0 , 0 ) = 0 f_y(0,0)=0 f y ( 0 , 0 ) = 0 Why? Plug y = 0 y=0 y = 0 0 0 0 But continuity fails: approach along y = x y=x y = x f ( x , x ) = x 2 2 x 2 = 1 2 ≠ 0 = f ( 0 , 0 ) f(x,x)=\frac{x^2}{2x^2}=\tfrac12\neq 0=f(0,0) f ( x , x ) = 2 x 2 x 2 = 2 1 = 0 = f ( 0 , 0 ) Why this kills it? Differentiable ⇒ continuous; it isn't continuous, so it can't be differentiable — even though both partials exist!Moral: partials = 0 is a mirage of niceness. The diagonal direction reveals the truth.
Common mistake Steel-man: "Partials exist ⇒ differentiable"Why it feels right: In 1D, "derivative exists" is differentiability — so we expect the same. And f x , f y f_x,f_y f x , f y Why it's wrong: Partials only test the two axis directions . A function can be smooth along x x x y y y diagonal (Example 2). Differentiability demands good behaviour from every direction simultaneously — a 2D requirement.
The fix: Either verify the o ( ∥ h ∥ ) o(\|\mathbf h\|) o ( ∥ h ∥ ) or check the partials are continuous (C 1 C^1 C 1 C 1 C^1 C 1
Common mistake Confusing "directional derivatives exist in all directions" with differentiabilityEven if directional derivatives exist along every line through a point, f f f D u f = ∇ f ⋅ u D_{\mathbf u}f=\nabla f\cdot\mathbf u D u f = ∇ f ⋅ u
Recall Feynman: explain to a 12-year-oldImagine a bumpy hill. Standing at a spot, you check the slope walking exactly East and exactly North — those are the "partial slopes." But that tells you nothing about a sudden cliff in the North-East direction! A hill is "smooth" (differentiable) only if, no matter which way you step, the ground rises like a flat tilted board you placed under your feet. If even one diagonal direction betrays you, the hill is secretly broken at that point — even though East and North looked perfectly fine.
"Partials are blind to diagonals; only the plane sees all."
And the hierarchy: C ontinuously-diff → D ifferentiable → C ontinuous → (and separately) P artials.
"C ould D ucks C arry P lanes?" — left implies right, never the reverse for the loner P.
Define differentiability of f : R 2 → R f:\mathbb R^2\to\mathbb R f : R 2 → R a \mathbf a a lim h → 0 ∣ f ( a + h ) − f ( a ) − ∇ f ( a ) ⋅ h ∣ ∥ h ∥ = 0 \lim_{\mathbf h\to 0}\frac{|f(\mathbf a+\mathbf h)-f(\mathbf a)-\nabla f(\mathbf a)\cdot\mathbf h|}{\|\mathbf h\|}=0 lim h → 0 ∥ h ∥ ∣ f ( a + h ) − f ( a ) − ∇ f ( a ) ⋅ h ∣ = 0 .
Why isn't "partials exist" enough for differentiability? Partials test only the axis directions; the function can be discontinuous/jagged along diagonals (e.g.
x y / ( x 2 + y 2 ) xy/(x^2+y^2) x y / ( x 2 + y 2 ) ).
What is the sufficient C 1 C^1 C 1 If
f x , f y f_x,f_y f x , f y exist and are continuous near
a \mathbf a a , then
f f f is differentiable at
a \mathbf a a .
Differentiable implies what weaker property? Continuity (and existence of all directional derivatives).
What must the linear map equal if f f f The gradient
∇ f = ( f x , f y ) \nabla f=(f_x,f_y) ∇ f = ( f x , f y ) — found by approaching along each axis.
Why divide the error by ∥ h ∥ \|\mathbf h\| ∥ h ∥ To force the error to be higher-order (
o ( ∥ h ∥ ) o(\|\mathbf h\|) o ( ∥ h ∥ ) ), making it a genuine tangent plane not just any touching plane.
For f ( x , y ) = x y x 2 + y 2 f(x,y)=\frac{xy}{x^2+y^2} f ( x , y ) = x 2 + y 2 x y f x ( 0 , 0 ) , f y ( 0 , 0 ) f_x(0,0),f_y(0,0) f x ( 0 , 0 ) , f y ( 0 , 0 ) f f f Both
= 0 =0 = 0 , but
f f f is not even continuous (limit
1 2 \tfrac12 2 1 along
y = x y=x y = x ), so not differentiable.
Linearization formula for estimation f ( a + h ) ≈ f ( a ) + ∇ f ( a ) ⋅ h f(\mathbf a+\mathbf h)\approx f(\mathbf a)+\nabla f(\mathbf a)\cdot\mathbf h f ( a + h ) ≈ f ( a ) + ∇ f ( a ) ⋅ h .
Limit definition: error over norm h to 0
Gradient nabla f = fx, fy
Higher-order error o of norm h
Trap: partials alone insufficient
Intuition Hinglish mein samjhoDekho, single-variable calculus mein differentiable hone ka matlab tha ek tangent line ka hona. Multivariable mein yeh ban jata hai tangent plane . Ab sabse bada confusion yahi hai: students sochte hain ki agar partial derivatives f x f_x f x f y f_y f y Galat! Partial derivatives sirf do directions — x-axis aur y-axis — ko check karte hain. Diagonal direction mein function phat sakta hai, aur partials ko pata bhi nahi chalega.
Asli definition yeh kehti hai: error term f ( a + h ) − f ( a ) − ∇ f ⋅ h f(\mathbf a+\mathbf h)-f(\mathbf a)-\nabla f\cdot \mathbf h f ( a + h ) − f ( a ) − ∇ f ⋅ h ∥ h ∥ \|\mathbf h\| ∥ h ∥ har direction se zero pe jana chahiye. Yeh ∥ h ∥ \|\mathbf h\| ∥ h ∥ f = x y / ( x 2 + y 2 ) f=xy/(x^2+y^2) f = x y / ( x 2 + y 2 ) y = x y=x y = x 1 / 2 1/2 1/2
Practical trick (yeh 80/20 hai): raw limit har baar check karne ki zaroorat nahi. Agar f x f_x f x f y f_y f y continuous hain point ke aas-paas, toh function automatically differentiable hai — isko C 1 C^1 C 1 sin \sin sin exp \exp exp C 1 C^1 C 1 C 1 ⇒ C^1 \Rightarrow C 1 ⇒ ⇒ \Rightarrow ⇒
