Differential Calculus: Several Variables (9hrs) · Engineering Mathematics - I

differentiability of functions of Several Variables

Watch lectures

differentiability of functions of Several Variables

1. From Continuity to Differentiability — The Hierarchy

For f(x,y)f(x,y) at (a,b)(a,b):

Differentiable⇒Continuous⇒LimitexistsDifferentiable ⇒ Continuous ⇒ Limit exists

but converses fail. ff can be continuous without partials; partials can exist without continuity; and even both partials existing does not guarantee differentiability — you need a stronger, linear approximation condition.

2. Definition — Differentiable at (a,b)(a,b)

f(x,y)f(x,y) is differentiable at (a,b)(a,b) if there are numbers A=fx(a,b),B=fy(a,b)A = f_x(a,b), B = f_y(a,b) such that:

f(a+h,b+k)−f(a,b)=Ah+Bk+o(√(h2+k2))f(a+h, b+k) - f(a,b) = A h + B k + o(√(h²+k²))

i.e. the error after the plane approximation goes to zero faster than the distance. Equivalently:

lim(h,k)→(0,0)[f(a+h,b+k)−f(a,b)−Ah−Bk]/√(h2+k2)=0lim (h,k)→(0,0) [f(a+h,b+k)-f(a,b)-A h -B k]/√(h²+k²) = 0

Geometrically: surface has a non-vertical tangent plane at PP:

z−f(a,b)=fx(a,b)(x−a)+fy(a,b)(y−b)z - f(a,b) = f_x(a,b)(x-a) + f_y(a,b)(y-b)

and ff hugs that plane to first order.

3. Sufficient Condition — The One You Will Use

If fxf_x and fyf_y exist and are continuous in a neighbourhood of (a,b)(a,b), then ff is differentiable at (a,b)(a,b) (and hence continuous). This is the practical test — check continuity of partials, not theε-δ definition.

Continuity of f_x, f_y near (a,b)  ⇒  differentiability at (a,b)  ⇒  continuity

4. Differentials and Total Derivative

  • Total differential:
dz=fxdx+fydydz = f_x dx + f_y dy

is the linear part of the increment ΔfΔf.

  • If x=x(t),y=y(t)x=x(t), y=y(t), then:
df/dt=fxdx/dt+fydy/dtdf/dt = f_x dx/dt + f_y dy/dt
  • If x=x(u,v),y=y(u,v)x=x(u,v), y=y(u,v), then fu=fxxu+fyyuf_u = f_x x_u + f_y y_u (chain rule follows from differentiability).

5. Counter-Example — Why Partials Alone Are Not Enough

f(x,y)= x y/(x²+y²) for (x,y)≠(0,0), f(0,0)=0:
 f_x(0,0)= lim h→0 [f(h,0)-0]/h =0 ; similarly f_y(0,0)=0 (both exist)
 But f is not continuous at (0,0) (limit does not exist) ⇒ cannot be differentiable.

Stronger trap:

f(x,y)=√(x2+y2)sin(1/√(x2+y2))f(x,y)= √(x²+y²) sin(1/√(x²+y²))

is continuous but not differentiable at 0 (no plane).

Another classic:

f(x,y)=∣x∣+∣y∣f(x,y)= |x|+|y|

has partials nowhere? At (0,0) partials do not exist.

6. Worked Examples

(a) f(x,y)= x^2 y + y sin x : f_x=2x y + y cos x, f_y= x^2 + sin x ; both continuous everywhere
    ⇒ differentiable everywhere; at (1,0): dz = (0)dx + (1)dy ; tangent plane z = (x-1)·0 + y·1 ✓

(b) f(x,y)= x^3+y^3 at (0,0): f_x=3x^2, f_y=3y^2 continuous ⇒ differentiable; plane z=0 ✓

(c) Check differentiability at (0,0) for f= x^2 y/(x^2+y^2) , f(0,0)=0:
    f_x(0,0)=0, f_y(0,0)=0 ; check limit [f(h,k)-0]/√(h²+k²)= h^2 k/(h^2+k^2)^{3/2}
    |h^2 k| ≤ (h^2+k^2)^{3/2}/(3√3)? Actually bounded by r|cos^2 sin|→0 ⇒ differentiable ✓

(d) f(x,y)= (x y^2)/(x^2+y^4), f(0,0)=0 : f_x(0,0)=0, f_y(0,0)=0 but [f(h,k)-0 -0]/r does not →0 along k^2=h ⇒ not differentiable ✓

7. Exam Points

  • State definition of differentiability and tangent plane — 3-5 marks, frequent.
  • Show fx,fyf_x, f_y continuous ⇒ differentiable ⇒ continuous — 5 marks, frequent.
  • Prove ff not differentiable by showing discontinuity or limit in definition fails — 5-7 marks, very frequent.
  • Find dzdz and df/dtdf/dt via chain rule — 5 marks, frequent.
  • Give example where partials exist but ff not differentiable / not continuous — 5 marks, almost certain.
Official AppAndroid App

PYQDeck for Android

Solve BEU previous-year exam papers anywhere, anytime. Fast search, offline paper viewing, and step-by-step solutions designed specifically for your phone.

← All Engineering Mathematics - I modules