\[f'(a) = \lim_{{h \to 0}} \frac{f(a + h) - f(a)}{h}.\]

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

- 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 \(x = a\).

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 \(f'(x) = g'(u(x)) \cdot u'(x)\).

- \(\frac{d}{dx}(\cos(x)) = -\sin(x)\)

- \(\frac{d}{dx}(\tan(x)) = \sec^2(x)\)

- \(\frac{d}{dx}(e^x) = e^x\)

- \(\frac{d}{dx}(\ln(x)) = \frac{1}{x}\)

If a function is differentiable at a point \(a\), it must be continuous at \(a\).

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 \(x = 0\)).

Check if the function \( f(x) = |x| \) is differentiable at \( x = 0 \).

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

Prove that the function \( f(x) = x^3 \) is differentiable at every point and find \( f'(2) \).

The derivative \( f'(x) = 3x^2 \) is defined for all real numbers, so \( f(x) \) is differentiable everywhere.

At \( x = 2 \), \( f'(2) = 3 \times 2^2 = 12 \).

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 \( x = 3 \).

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

If \( f(x) = \sqrt{3x + 2} \), find \( f'(1) \).

\( f'(x) = \frac{3}{2\sqrt{3x + 2}} \).

So, \( f'(1) = \frac{3}{2\sqrt{5}} \).

**Notes**-
**NCERT Solutions & Assignments**- NCERT Solution for Continuity And differentiability Exercise 5.1
- NCERT Solution for Continuity And differentiability Exercise 5.2
- NCERT Solution for Continuity And differentiability Exercise 5.3
- NCERT Solution for Continuity And differentiability Exercise 5.4
- Continuity and Differentiability class 12 Important questions

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