Geometric Interpretation of Derivatives

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• 1. Definition of Geometric Interpretation of Derivatives
• 2. Explanation & Intuition
• 3. Solved Examples
• 4. Conclusion

Geometric Interpretation of Derivatives:
The derivative of a function at a point represents the slope of the tangent line to the curve at that point. Geometrically, it measures how steep the function is at any given point.

The derivative at a point $x = a$ is given by:

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

This formula gives the slope of the tangent line at $x = a$ by finding the rate of change of the function over a very small interval around $x = a$.

Example 1: Differentiating a Polynomial

Differentiate $f(x) = x^3 - 4x^2 + 2x + 1$.

Solution:

The derivative is:

$f'(x) = \dfrac{d}{dx} \left( x^3 - 4x^2 + 2x + 1 \right)$

Using the power rule:

$f'(x) = 3x^2 - 8x + 2$

At $x = 1$, the derivative is:

$f'(1) = 3(1)^2 - 8(1) + 2 = 3 - 8 + 2 = -3$

The slope of the tangent line at $x = 1$ is -3, indicating that the function is decreasing at this point.

Thus, the final answer is:

$\boxed{f'(1) = -3}$.

Example 2: Differentiating a Trigonometric Function

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

Solution:

The derivative is:

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

Using the chain rule:

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

At $x = \frac{\pi}{4}$, the derivative is:

$f' \left( \frac{\pi}{4} \right) = 2 \cos\left( 2 \cdot \frac{\pi}{4} \right) - \sin\left( \frac{\pi}{4} \right)$

$f' \left( \frac{\pi}{4} \right) = 2 \cos\left( \frac{\pi}{2} \right) - \sin\left( \frac{\pi}{4} \right)$

$f' \left( \frac{\pi}{4} \right) = 0 - \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2}$

Thus, the final answer is:

$\boxed{f' \left( \frac{\pi}{4} \right) = -\frac{\sqrt{2}}{2}}$.

Example 3: Applying Derivative to Motion

The position of an object moving along a straight line is given by:

$s(t) = 4t^3 - 3t^2 + 2t$

Find the velocity of the object at $t = 2$.

Solution:

The velocity is the derivative of the position function:

$v(t) = \dfrac{d}{dt} \left( 4t^3 - 3t^2 + 2t \right)$

Differentiating:

$v(t) = 12t^2 - 6t + 2$

At $t = 2$, the velocity is:

$v(2) = 12(2)^2 - 6(2) + 2 = 12(4) - 12 + 2 = 48 - 12 + 2 = 38$

Thus, the velocity of the object at $t = 2$ is:

$\boxed{v(2) = 38} \, \text{m/s}$.

Example 4: Derivative of an Exponential Function

Differentiate $f(x) = e^{3x^2}$.

Solution:

Using the chain rule, let:

• Outer function: $f(u) = e^u$, so $f'(u) = e^u$.
• Inner function: $g(x) = 3x^2$, so $g'(x) = 6x$.

Applying the chain rule:

$f'(x) = e^{3x^2} \cdot 6x$

Thus, the final answer is:

$\boxed{f'(x) = 6x e^{3x^2}}$.