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