Higher-Order Differentiation in Calculus

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• 1. Definition of Higher Order Derivatives
• 2. Formula Used
• 3. Solved Examples
• 4. Conclusion

Higher Order Derivatives:
The higher order derivatives are the derivatives of the first derivative of a function. These derivatives give us more information about the function, such as concavity and rate of change of the slope.

If $f(x)$ is a function, its first derivative is denoted as:

$f'(x) = \dfrac{d}{dx} \bigg(f(x)\bigg)$

The second derivative is the derivative of the first derivative, denoted as:

$f''(x) = \dfrac{d^2}{dx^2} \bigg(f(x)\bigg)$

Similarly, higher order derivatives are denoted as:

$f^{(n)}(x) = \dfrac{d^n}{dx^n} \bigg(f(x)\bigg)$

Where $n$ represents the order of the derivative.

Example 1: Second Derivative of a Polynomial

Differentiate $f(x) = 3x^4 - 4x^3 + 2x^2 - x + 5$ up to the second derivative.

Solution:

First, find the first derivative:

$f'(x) = \dfrac{d}{dx} (3x^4 - 4x^3 + 2x^2 - x + 5)$

$f'(x) = 12x^3 - 12x^2 + 4x - 1$

Now, differentiate again to find the second derivative:

$f''(x) = \dfrac{d}{dx} (12x^3 - 12x^2 + 4x - 1)$

$f''(x) = 36x^2 - 24x + 4$

Thus, the second derivative is:

$\boxed{f''(x) = 36x^2 - 24x + 4}$.

Example 2: Finding Acceleration in Physics

Given the position function of an object $s(t) = 4t^3 - 3t^2 + 2t$, find the velocity and acceleration.

Solution:

First, differentiate to find velocity:

$v(t) = \dfrac{ds}{dt} = 12t^2 - 6t + 2$

Now, differentiate to find acceleration:

$a(t) = \dfrac{dv}{dt} = 24t - 6$

Thus, the acceleration is:

$\boxed{a(t) = 24t - 6}$

This shows how higher-order derivatives (velocity and acceleration) describe the motion of an object.

Example 3: Analyzing Concavity and Inflection Points

Find the second derivative of $f(x) = \ln(x^2 + 1)$ and determine concavity.

Solution:

First, differentiate to find the first derivative:

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

Now, differentiate again to find the second derivative:

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

Applying the quotient rule:

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

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

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

Thus, the second derivative is:

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

To find the concavity, we analyze the sign of $f''(x)$. The function is concave up where $f''(x) > 0$ and concave down where $f''(x) < 0$.

Example 4: Second Derivative of a Logarithmic Function

Differentiate $f(x) = \ln(x^3 + 2x)$ up to the second derivative.

Solution:

First, differentiate to find the first derivative:

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

Now, differentiate again to find the second derivative:

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

This is a complex application of the quotient rule, but after applying it, we get:

$f''(x) = \dfrac{(x^3 + 2x)(6x) - (3x^2 + 2)(3x^2 + 2)}{(x^3 + 2x)^2}$