# Differentiability Notes | Class 12 Maths

## Definition of Differentiability

A function $f(x)$ is said to be differentiable at a point $a$ if the limit exists:
$f'(a) = \lim_{{h \to 0}} \frac{f(a + h) - f(a)}{h}.$

This limit is the derivative of $f$ at the point $a$, denoted as $f'(a)$. If this limit exists, the function is differentiable at $a$.

## Geometric Interpretation of the Derivative

- The derivative $f'(a)$ represents the slope of the tangent line to the graph of $f$ at the point $(a, f(a))$.
- If $f'(a) > 0$, the graph of $f$ is increasing at $x = a$.
- If $f'(a) < 0$, the graph of $f$ is decreasing at $x = a$.
- If $f'(a) = 0$, the graph of $f$ has a horizontal tangent at $x = a$.

## Rules for Finding Derivatives

1. Constant Rule
The derivative of a constant $c$ is zero: $d/dx(c) = 0$.

2. Power Rule
If $f(x) = x^n$, where $n$ is a constant, then $f'(x) = nx^{n-1}$.

3. Sum/Difference Rule If $f(x) = g(x) \pm h(x)$, then $f'(x) = g'(x) \pm h'(x)$.

4. Product Rule
If $f(x) = g(x) \cdot h(x)$, then $f'(x) = g(x)h'(x) + g'(x)h(x)$.

5. Quotient Rule
If $f(x) = \frac{g(x)}{h(x)}$, then $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$.

6.Chain Rule
If $f(x) = g(u(x))$, where $u(x)$ is a function of $x$, then $f'(x) = g'(u(x)) \cdot u'(x)$.

## Common Derivatives

- $\frac{d}{dx}(\sin(x)) = \cos(x)$
- $\frac{d}{dx}(\cos(x)) = -\sin(x)$
- $\frac{d}{dx}(\tan(x)) = \sec^2(x)$
- $\frac{d}{dx}(e^x) = e^x$
- $\frac{d}{dx}(\ln(x)) = \frac{1}{x}$

## Differentiability in a interval

A function is said to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b]. As in case of continuity, at the end points a and b, we take the right hand limit and left hand limit, which are nothing but left hand derivative and right hand derivative of the function at a and b respectively.

## Differentiability and Continuity

Theorem 1
If a function is differentiable at a point $a$, it must be continuous at $a$.
Proof $f'(a) = \lim_{{x \to a}} \frac{f(x) - f(a)}{x-a}.$
Now for $x \ne a$
$f(x) - f(a)=\frac {f(x) - f(a)}{x-a} (x-a)$
$\lim_{{x \to a}} [f(x) - f(a)]= \lim_{{x \to a}} [\frac {f(x) - f(a)}{x-a} (x-a)$
$\lim_{{x \to a}} f(x) - \lim_{{x \to a}} f(a) = \lim_{{x \to a}} \frac {f(x) - f(a)}{x-a} \lim_{{x \to a}} (x-a)$
$\lim_{{x \to a}} f(x) - f(a) = f'(a). 0=0$
or $\lim_{{x \to a}} f(x= f(a)$
Hence continous

However, a function can be continuous at a point without being differentiable at that point (e.g., the absolute value function at $x = 0$).

## Solved Examples

Question 1
Check if the function $f(x) = |x|$ is differentiable at $x = 0$.
Solution It is known that a function $f$ is differentiable at a point $x=c$ in its domain if the right hand limit and the left hand limit are finite and equal.
To check the differentiability of the given function at x=1,
The right hand and the left hand limits where x=c are
$\lim_{h \to 0^{+}}\frac{f(c+h)-f(c)}{h}$ and $\lim_{h \to 0^{-}}\frac{f(c+h)-f(c)}{h}$
Considering the right hand limit of the given function at x=0
$\lim_{h \to 0^{+}}\frac{f(0+h)-f(0)}{h}$
$=\lim_{h \to 0^{+}}\frac{h}{h}$
$=1$
Considering the left hand limit of the given function at x=0
$\lim_{h \to 0^{-}}\frac{f(0+h)-f(0)}{h}$
$=\lim_{h \to 0^{-}}\frac{-h}{h}$
$=-1$
Since the left and right hand limits of f at x = 0 are not equal, f is not differentiable at x = 0

Question 2
Prove that the function $f(x) = x^3$ is differentiable at every point and find $f'(2)$.
Solution
The derivative $f'(x) = 3x^2$ is defined for all real numbers, so $f(x)$ is differentiable everywhere.
At $x = 2$, $f'(2) = 3 \times 2^2 = 12$.

Question 3
Check the differentiability of $f(x) = \begin{cases} x^2 + 2x, & \text{if } x < 3 \\ 3x + 7, & \text{if } x \geq 3 \end{cases}$ at $x = 3$.
Solution
LHD at $x = 3$
$\lim_{h \to 0^{-}}\frac{f(3+h)-f(3)}{h}$
$=\lim_{h \to 0^{-}}\frac{ (3+h)^2 + 2(3+h) - 3^2 -2 \times 3}{h}$
$=\lim_{h \to 0^{-}}\frac{h^2 +6h + 2h}{h}$
$=\lim_{h \to 0^{-}} (h +8) =8$
RHD at $x = 3$
$\lim_{h \to 0^{+}}\frac{f(3+h)-f(3)}{h}$
$=\lim_{h \to 0^{+}}\frac{ 3(h+3) + 7 -3 \times 3 -7}{h}$
$=\lim_{h \to 0^{+}}\frac{3h}{h}$
$=\lim_{h \to 0^{+}} (3) =3$
Since the left and right hand limits of f at x = 3 are not equal, f is not differentiable at x = 3

Question 4
If $f(x) = \sqrt{3x + 2}$, find $f'(1)$.
Solution
$f'(x) = \frac{3}{2\sqrt{3x + 2}}$.
So, $f'(1) = \frac{3}{2\sqrt{5}}$.

Go back to Class 12 Main Page using below links