Definition of Derivative in Calculus

Neetesh Kumar | February 06, 2025 $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space$ Share this Page on:

• 1. Definition of Derivative
• 2. Different Notations for Derivatives
• 3. Limit Definition of Derivatives
• 4. Condition of Differentiability
• 5. Relationship between Continuity & Differentiability
• 6. Derivative Examples with Step-by-Step Solutions
• 7. Conclusion

The derivative of a function $f(x)$ with respect to $x$ is denoted as $f'(x)$ and is defined as: $\boxed{\lim\limits_{h \to 0} \bigg(\dfrac{f(x+h) - f(x)}{h}\bigg)}$

The limit definition of a derivative is the same as the definition of the derivative. : $\boxed{f'(x) =\lim\limits_{h \to 0} \bigg(\dfrac{f(x+h) - f(x)}{h}\bigg)}$

• A function $f(x)$ is said to be differentiable at $x = a$ if $f'(a)$ exists and $f(x)$ is called differentiable on an interval if the derivative exists for each point in that interval.

• A function $f(x)$ is differentiable at $x = a$ if and only if the following limit exists: $\lim\limits_{h \to 0} \bigg(\dfrac{f(a+h) - f(a)}{h}\bigg)$. If this limit exists, then $f(x)$ is differentiable at $x=a$, and the derivative $f'(a)$ gives the slope of the tangent line at $x=a$.

• A function $f(x)$ is differentiable on an open interval (a, b) then it means $f(x)$ is differentiable at every point in that interval.

A function $f(x)$ is said to be differentiable at $x = a$ then we can say that $f(x)$ is continuous at $x = a$ but vice-versa is not true.

Example 1:

Given $f(x) = x^2$, find $f'(x)$ and $f'(2)$ using the limit definition of derivative.

Solution:

We know that $f(x+h) = (x+h)^2$.

By using the limit definition of derivative:

$f'(x) =\lim\limits_{h \to 0} \bigg(\dfrac{f(x+h) - f(x)}{h}\bigg) = \lim\limits_{h \to 0} \bigg(\dfrac{(x+h)^2 - x^2}{h}\bigg)$

Expanding $(x+h)^2$ term:

$f'(x) = \lim\limits_{h \to 0} \bigg(\dfrac{x^2 + 2xh + h^2 - x^2}{h}\bigg) = \dfrac{2xh + h^2}{h} = 2x + h$

Taking the limit:

$f'(x) = \lim\limits_{h \to 0} (2x+h) = 2x$

Thus,

$f'(2) = 2(2) = 4$

Example 2:

Given $f(x) = \sqrt{x}$, find $f'(x)$ and $f'(8)$ using the limit definition of derivative.

Solution:

We know that $f(x+h) = \sqrt{(x+h)}$.

By using the limit definition of derivative:

$f'(x) =\lim\limits_{h \to 0} \bigg(\dfrac{f(x+h) - f(x)}{h}\bigg) = \lim\limits_{h \to 0} \bigg(\dfrac{\sqrt{x+h} - \sqrt{x}}{h}\bigg)$

To simplify, multiply the numerator and denominator by the conjugate:

$f'(x) = \lim\limits_{h \to 0} \bigg(\dfrac{(\sqrt{x+h} - \sqrt{x}) (\sqrt{x+h} + \sqrt{x})}{h (\sqrt{x+h} + \sqrt{x})}\bigg)$

Since $(a - b)(a + b) = a^2 - b^2$, we get:

$f'(x) = \lim\limits_{h \to 0} \bigg(\dfrac{(x+h) - x}{h (\sqrt{x+h} + \sqrt{x})}\bigg) = \lim\limits_{h \to 0} \bigg(\dfrac{h}{h (\sqrt{x+h} + \sqrt{x})}\bigg)$

Canceling $h$:

$f'(x) = \lim\limits_{h \to 0} \bigg(\dfrac{1}{\sqrt{x+h} + \sqrt{x}}\bigg)$

Taking the limit as $h \to 0$:

$f'(x) = \dfrac{1}{2\sqrt{x}}$

Thus,

$f'(8) = \dfrac{1}{2\sqrt{8}} = \dfrac{1}{4}$

Example 3:

Given $f(x) = 2x^2 - 5x$, find $f'(x)$ using the limit definition of derivative.

Solution:

We know that $f(x+h) = 2(x+h)^2 - 5(x+h)$.

By using the limit definition of derivative:

$f'(x) =\lim\limits_{h \to 0} \bigg(\dfrac{f(x+h) - f(x)}{h}\bigg)$

Substituting $f(x+h)$ and $f(x)$:

$f'(x) = \lim\limits_{h \to 0} \bigg(\dfrac{2(x+h)^2 - 5(x+h) - (2x^2 - 5x)}{h}\bigg)$

Expanding $(x+h)^2$:

$(x+h)^2 = x^2 + 2xh + h^2$

Substituting this back:

$f'(x) = \lim\limits_{h \to 0} \bigg(\dfrac{2(x^2 + 2xh + h^2) - 5x - 5h - 2x^2 + 5x}{h}\bigg)$

Distribute $2$:

$f'(x) = \lim\limits_{h \to 0} \bigg(\dfrac{2x^2 + 4xh + 2h^2 - 5x - 5h - 2x^2 + 5x}{h}\bigg)$

Canceling like terms ($2x^2 - 2x^2$ and $-5x + 5x$):

$f'(x) = \lim\limits_{h \to 0} \bigg(\dfrac{4xh + 2h^2 - 5h}{h}\bigg)$

Factor out $h$ in the numerator:

$f'(x) = \lim\limits_{h \to 0} \bigg(\dfrac{h(4x + 2h - 5)}{h}\bigg)$

Canceling $h$:

$f'(x) = \lim\limits_{h \to 0} (4x + 2h - 5)$

Taking the limit as $h \to 0$:

$f'(x) = 4x - 5$

Thus, the derivative of $f(x)$ is:

$f'(x) = 4x - 5$

Example 4:

Estimate $f'(2)$ and $f'(2.5)$ if:

Solution:

We have that:

$f'(x) = \lim\limits_{h \to 0} \bigg(\dfrac{f(x+h) - f(x)}{h}\bigg)$

But if we take $h = 0.5$, we will obtain the following approximation:

$f'(x) \approx \dfrac{f(x+0.5) - f(x)}{0.5} = 2(f(x + 0.5) - f(x))$.

Thus,

$f'(2) \approx 2(f(2 + 0.5) - f(2)) = 2(f(2.5) - f(2)) = 2(6 - 5) = 2$.

$f'(2.5) \approx 2(f(2.5 + 0.5) - f(2.5)) = 2(f(3) - f(2.5)) = 2(10 - 6) = 8$.

Example 5:

Given $g(t) = \dfrac{t}{t+1}$, find $g'(t)$ using the limit definition of derivative.

Solution:

We know that:

$g(t+h) = \bigg(\dfrac{t+h}{(t+h) + 1}\bigg) = \bigg(\dfrac{t+h}{t+h+1}\bigg)$.

By using the limit definition of derivative:

$g'(t) = \lim\limits_{h \to 0} \bigg(\dfrac{g(t+h) - g(t)}{h}\bigg) = \lim\limits_{h \to 0} \bigg(\dfrac{\frac{t+h}{t+h+1} - \frac{t}{t+1}}{h}\bigg)$.

To simplify, find a common denominator:

$g'(t) = \lim\limits_{h \to 0} \bigg(\dfrac{\frac{(t+h)(t+1) - t(t+h+1)}{(t+h+1)(t+1)}}{h}\bigg)$.

Expanding the numerator:

$(t+h)(t+1) = t^2 + t + ht + h$,

$t(t+h+1) = t^2 + ht + t$.

Thus, the numerator becomes:

$t^2 + t + ht + h - (t^2 + ht + t) = h$.

Now, we simplify:

$g'(t) = \lim\limits_{h \to 0} \bigg(\dfrac{\frac{h}{(t+h+1)(t+1)}}{h}\bigg)$.

Canceling $h$:

$g'(t) = \lim\limits_{h \to 0} \dfrac{1}{(t+h+1)(t+1)}$.

Taking the limit as $h \to 0$:

$g'(t) = \dfrac{1}{(t+1)(t+1)} = \dfrac{1}{(t+1)^2}$.

Thus, the derivative of $g(t)$ is:

$g'(t) = \dfrac{1}{(t+1)^2}$.

Example 6:

Given $R(z) = (2z+5)$, find $R'(z)$ using the limit definition of derivative.

Solution:

We know that $R(z+h) = 2(z+h) + 5$.

By using the limit definition of derivative:

$R'(z) = \lim\limits_{h \to 0} \bigg(\dfrac{R(z+h) - R(z)}{h}\bigg)$

Substituting $R(z+h)$ and $R(z)$:

$R'(z) = \lim\limits_{h \to 0} \bigg(\dfrac{(2(z+h) + 5) - (2z+5)}{h}\bigg)$

Expanding:

$R'(z) = \lim\limits_{h \to 0} \bigg(\dfrac{(2z + 2h + 5) - (2z + 5)}{h}\bigg)$

Canceling like terms ($2z - 2z$ and $5 - 5$):

$R'(z) = \lim\limits_{h \to 0} \bigg(\dfrac{2h}{h}\bigg)$

Canceling $h$:

$R'(z) = \lim\limits_{h \to 0} 2$

Taking the limit as $h \to 0$:

$R'(z) = 2$

Thus, the derivative of $R(z)$ is:

$R'(z) = 2$

Example 7:

Given $f(x) = |x|$, find $f'(0)$ using the limit definition of the derivative.

Solution:

We use the limit definition of the derivative:

$f'(x) = \lim\limits_{h \to 0} \bigg(\dfrac{f(x+h) - f(x)}{h}\bigg)$

Substituting $x = 0$:

$f'(0) = \lim\limits_{h \to 0} \bigg(\dfrac{|0+h| - |0|}{h}\bigg)$

Since $|0| = 0$, this simplifies to:

$f'(0) = \lim\limits_{h \to 0} \bigg(\dfrac{|h|}{h}\bigg)$

Now, we analyze the left-hand limit (as $h \to 0^-$) and the right-hand limit (as $h \to 0^+$):

• For $h > 0$ (right-hand limit), $|h| = h$, so:

  $\lim\limits_{h \to 0^+} \frac{h}{h} = 1$.

• For $h < 0$ (left-hand limit), $|h| = -h$, so:

  $\lim\limits_{h \to 0^-} \frac{-h}{h} = -1$.

Since the left-hand limit ($-1$) and right-hand limit ($1$) are not equal, the derivative does not exist at $x = 0$.

Thus,

$f'(0)$ does not exist because the function $f(x) = |x|$ is not differentiable at $x = 0$.