Graphical Aspect of Derivatives

Neetesh Kumar | February 08, 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 Graphical Aspect of Derivatives
• 2. Formula and Graphical Intuition
• 3. Solved Examples
• 4. Conclusion

The graphical aspect of derivatives refers to the relationship between the derivative of a function and the slope of the tangent line at any point on the function's graph.

• If the derivative of a function at a point is positive, the function is increasing at that point.
• If the derivative is negative, the function is decreasing at that point.
• If the derivative is zero, the function has a horizontal tangent line, which indicates a potential maximum, minimum, or inflection point.

Mathematically, the derivative at any point $x = a$ is defined as:

$\boxed{f'(a) = \lim_{h \to 0} \bigg(\frac{f(a+h) - f(a)}{h}\bigg)}$

The derivative represents the slope of the tangent line to the curve at a point.

Example 1: Differentiating a Polynomial Function

Differentiate $f(x) = x^3 - 3x^2 + 2x - 1$.

Solution:

First, we differentiate each term:

$f'(x) = \dfrac{d}{dx} (x^3) - \dfrac{d}{dx} (3x^2) + \dfrac{d}{dx} (2x) - \dfrac{d}{dx} (1)$

$f'(x) = 3x^2 - 6x + 2$

This means that at any point on the curve, the slope of the tangent line is given by $3x^2 - 6x + 2$.

To find the slope at a specific point, substitute $x = 1$:

$f'(1) = 3(1)^2 - 6(1) + 2 = 3 - 6 + 2 = -1$

Thus, the slope of the tangent line at $x = 1$ is $\boxed{-1}$.

Example 2: Differentiating a Trigonometric Function

Differentiate $f(x) = \cos(2x)$.

Solution:

Using the chain rule:

$f'(x) = \dfrac{d}{dx} \left[ \cos(2x) \right]$

Using the derivative of cosine and chain rule:

$f'(x) = -\sin(2x) \cdot \dfrac{d}{dx} (2x)$

$f'(x) = -2\sin(2x)$

Thus, the derivative of $f(x) = \cos(2x)$ is: $\boxed{-2\sin(2x)}$.

To find the slope at a specific point, substitute $x = 0$: $f'(0) = -2\sin(0) = 0$

Thus, the slope of the tangent line at $x = 0$ is $\boxed{0}$.

Example 3: Finding the Slope of a Logarithmic Function

Differentiate $f(x) = \ln(4x^2 + 1)$.

Solution:

Using the chain rule:

$f'(x) = \dfrac{d}{dx} \left[ \ln(4x^2 + 1) \right]$

The derivative of the natural logarithm is:

$f'(x) = \dfrac{1}{4x^2 + 1} \cdot \dfrac{d}{dx} (4x^2 + 1)$

$f'(x) = \dfrac{1}{4x^2 + 1} \cdot 8x$

Thus, the final answer is: $\boxed{\frac{8x}{4x^2 + 1}}$.

To find the slope at a specific point, substitute $x = 1$:

$f'(1) = \dfrac{8(1)}{4(1)^2 + 1} = \dfrac{8}{5}$

Thus, the slope of the tangent line at $x = 1$ is: $\boxed{\frac{8}{5}}$.

Example 4: Finding the Tangent Line to an Exponential Function

Find the equation of the tangent line to $f(x) = e^{2x}$ at $x = 0$.

Solution:

First, differentiate $f(x) = e^{2x}$:

$f'(x) = 2e^{2x}$

Now, find $f(0)$:

$f(0) = e^{2(0)} = 1$

The slope at $x = 0$ is:

$f'(0) = 2e^{2(0)} = 2$

Thus, the equation of the tangent line at $x = 0$ is given by:

$y - f(0) = f'(0)(x - 0)$

$y - 1 = 2(x - 0)$

Simplifying:

$y = 2x + 1$

Thus, the equation of the tangent line is: $\boxed{y = 2x + 1}$.