Differentiability

Hello, math enthusiasts! Welcome back to KnowledgeKnot! Today, we will explore differentiability, one of the most fundamental concepts in calculus. Be sure to share and happy learning!

Definition of Differentiability

Differentiability determines whether a function has a derivative at a particular point. A function f(x)f(x) is said to be **differentiable** at a point cc if the derivative f(c)f'(c) exists. This means:

f(c)=limh0f(c+h)f(c)hf'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}

The derivative represents the slope of the tangent line to the curve at that point. If the derivative exists for every point in an interval, the function is differentiable over that interval.

Necessary Conditions for Differentiability

A function must satisfy these conditions to be differentiable at a point x=cx=c:

  • Continuity: The function must be continuous at cc. Differentiability implies continuity, but continuity does not guarantee differentiability.
  • Left-hand and Right-hand Derivatives: The left-hand derivative and the right-hand derivative at cc must exist and be equal:
    limh0f(c+h)f(c)h=limh0+f(c+h)f(c)h\lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h} = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}

Points of Non-Differentiability

A function is **not differentiable** at certain points due to:

  • Sharp Corners: For example, the function f(x)=xf(x) = |x| has a sharp corner at x=0x=0.
  • Discontinuities: If a function is not continuous at a point, it cannot be differentiable there.
  • Vertical Tangents: When the slope of the tangent becomes infinite, as in f(x)=x1/3f(x) = x^{1/3} at x=0x=0.

Example Question

Problem:

Determine whether the function f(x)=xf(x) = |x| is differentiable at x=0x=0.

Solution:

  1. Find the left-hand derivative:
    f(x)=x=x for x<0f(x) = |x| = -x \text{ for } x < 0
    limh0f(0+h)f(0)h=limh0h0h=1\lim_{h \to 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h - 0}{h} = -1
  2. Find the right-hand derivative:
    f(x)=x=x for x>0f(x) = |x| = x \text{ for } x > 0
    limh0+f(0+h)f(0)h=limh0+h0h=1\lim_{h \to 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h - 0}{h} = 1
  3. Since the left-hand and right-hand derivatives are not equal, the function is not differentiable at x=0x=0.

Applications of Differentiability

Differentiability has wide-ranging applications in mathematics and beyond:

  • Physics: Calculating velocity and acceleration from position-time graphs.
  • Economics: Analyzing marginal costs and revenues.
  • Engineering: Designing systems with smooth transitions, such as in control theory.
  • Machine Learning: Optimization of functions during training.

Practice Question

Problem:

Verify the differentiability of the function f(x)={x2if x12x+1if x>1f(x) = \begin{cases} x^2 & \text{if } x \leq 1 \\ 2x + 1 & \text{if } x > 1 \end{cases} at x=1x=1.

Solution:

  1. Check if the function is continuous at x=1x=1:
    f(1)=12=1f(1) = 1^2 = 1
    For x1x \to 1^-:
    limx1f(x)=limx1x2=1\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = 1
    For x1+x \to 1^+:
    limx1+f(x)=limx1+(2x+1)=3\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x + 1) = 3
    Since the function is not continuous at x=1x=1, it is not differentiable.

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