Rules of Differentiation and Differentiability

1. Differentiability

A function $f(x)$ is said to be differentiable at a point $x=c$ if the derivative $f'(c)$ exists. For the derivative to exist, the limit in the first principle definition must exist. This means the limit from the left (Left-Hand Derivative) must equal the limit from the right (Right-Hand Derivative).

• Left-Hand Derivative (LHD): $$Lf'(c) = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h} = \lim_{h \to 0^+} \frac{f(c-h) - f(c)}{-h}$$

• Right-Hand Derivative (RHD): $$Rf'(c) = \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h}$$

A function is differentiable at $c$ if and only if LHD = RHD and both are finite.

Theorem: Differentiability implies Continuity

If a function is differentiable at a point, then it must be continuous at that point. However, the converse is not true. A function can be continuous at a point but not differentiable. A classic example is $f(x) = |x|$ at $x=0$, which is continuous but has a sharp corner, making it non-differentiable.

2. Derivatives of Standard Functions

• Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$
• Trigonometric Functions:
  • $\frac{d}{dx}(\sin x) = \cos x$
  • $\frac{d}{dx}(\cos x) = -\sin x$
  • $\frac{d}{dx}(\tan x) = \sec^2 x$
• Exponential & Logarithmic Functions:
  • $\frac{d}{dx}(e^x) = e^x$
  • $\frac{d}{dx}(a^x) = a^x \ln a$
  • $\frac{d}{dx}(\ln x) = \frac{1}{x}$

3. Algebra of Derivatives (Rules of Differentiation)

Instead of using the first principle every time, we use a set of rules to find derivatives quickly. Let $u = f(x)$ and $v = g(x)$ be differentiable functions.

1. Constant Rule: If $f(x) = c$ (a constant), then $f'(x) = 0$. $$\frac{d}{dx}(c) = 0$$

2. Power Rule: If $f(x) = x^n$ for any real number $n$, then $f'(x) = nx^{n-1}$. $$\frac{d}{dx}(x^n) = nx^{n-1}$$

3. Constant Multiple Rule: The derivative of a constant times a function is the constant times the derivative of the function. $$\frac{d}{dx}(c \cdot u) = c \cdot \frac{du}{dx}$$

4. Sum and Difference Rule: The derivative of a sum or difference of functions is the sum or difference of their derivatives. $$\frac{d}{dx}(u \pm v) = \frac{du}{dx} \pm \frac{dv}{dx}$$

5. Product Rule: $$\frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx}$$ In words: "The first function times the derivative of the second, plus the second function times the derivative of the first."

6. Quotient Rule: $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$ In words: "Low d-high minus high d-low, square the bottom and away we go."

7. Chain Rule: Used for differentiating composite functions (a function inside another function). If $y = f(g(x))$, let $u=g(x)$, so $y=f(u)$. Then: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ In words: "Differentiate the outer function (leaving the inside alone), then multiply by the derivative of the inner function."

Solved Examples

Example 1: Product Rule

Question: Find the derivative of $f(x) = x^2 \sin x$.

Solution: Let $u = x^2$ and $v = \sin x$. Then $u' = 2x$ and $v' = \cos x$. Using the product rule, $f'(x) = u v' + v u'$: $$f'(x) = (x^2)(\cos x) + (\sin x)(2x) = x^2 \cos x + 2x \sin x$$

Example 2: Quotient Rule

Question: Find the derivative of $f(x) = \frac{e^x}{x^3}$.

Solution: Let $u = e^x$ and $v = x^3$. Then $u' = e^x$ and $v' = 3x^2$. Using the quotient rule, $f'(x) = \frac{v u' - u v'}{v^2}$: $$f'(x) = \frac{(x^3)(e^x) - (e^x)(3x^2)}{(x^3)^2} = \frac{x^2e^x(x - 3)}{x^6} = \frac{e^x(x-3)}{x^4}$$

Example 3: Chain Rule

Question: Find the derivative of $y = \cos(x^2 + 1)$.

Solution: This is a composite function. The outer function is $\cos(u)$ and the inner function is $u = x^2 + 1$.

1. Derivative of the outer function: $\frac{d}{du}(\cos u) = -\sin u = -\sin(x^2+1)$.
2. Derivative of the inner function: $\frac{d}{dx}(x^2+1) = 2x$.
3. Multiply them together (Chain Rule): $$\frac{dy}{dx} = -\sin(x^2+1) \cdot (2x) = -2x \sin(x^2+1)$$

Example 4: Checking Differentiability

Question: Check if the function $f(x) = |x-2|$ is differentiable at $x=2$.

Solution: First, write the function in piecewise form: $f(x) = \begin{cases} x-2, & x \geq 2 \\ -(x-2), & x < 2 \end{cases}$

1. Right-Hand Derivative (RHD) at x=2: For $x>2$, $f(x)=x-2$, so the derivative is $f'(x)=1$. Thus, RHD = 1.

2. Left-Hand Derivative (LHD) at x=2: For $x<2$, $f(x)=-(x-2)=2-x$, so the derivative is $f'(x)=-1$. Thus, LHD = -1.

Since RHD (1) $\neq$ LHD (-1), the function is not differentiable at $x=2$. This is expected as the graph has a sharp corner at that point.