Successive Differentiation in Calculus

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• 1. Definition of Successive Differentiation
• 2. Formula Used
• 3. Solved Examples
• 4. Conclusion

Successive Differentiation:
Successive differentiation is the process of differentiating a function more than once, i.e., computing the first, second, third, and higher order derivatives of a function.

For example, if $f(x)$ is a differentiable function, then the first derivative is $f'(x)$, the second derivative is $f''(x)$, the third derivative is $f^{(3)}(x)$, and so on.

The n-th derivative of a function is denoted as:

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

Example 1: Differentiating a Polynomial Function

Differentiate $f(x) = 3x^4 - 5x^3 + 2x^2 - 7x + 1$ and find the second and third derivatives.

Solution:

First, find the first derivative:

$f'(x) = \dfrac{d}{dx} (3x^4 - 5x^3 + 2x^2 - 7x + 1)$

$f'(x) = 12x^3 - 15x^2 + 4x - 7$

Now, find the second derivative:

$f''(x) = \dfrac{d}{dx} (12x^3 - 15x^2 + 4x - 7)$

$f''(x) = 36x^2 - 30x + 4$

Finally, find the third derivative:

$f^{(3)}(x) = \dfrac{d}{dx} (36x^2 - 30x + 4)$

$f^{(3)}(x) = 72x - 30$

Thus, the derivatives are:

• First derivative: $f'(x) = 12x^3 - 15x^2 + 4x - 7$
• Second derivative: $f''(x) = 36x^2 - 30x + 4$
• Third derivative: $f^{(3)}(x) = 72x - 30$

Example 2: Differentiating a Trigonometric Function

Differentiate $f(x) = \sin(x^2)$ and find the first and second derivatives.

Solution:

First, apply the chain rule to find the first derivative:

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

$f'(x) = 2x \cos(x^2)$

Now, find the second derivative:

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

Using the product rule, we get:

$f''(x) = 2\cos(x^2) - 4x^2 \sin(x^2)$

Thus, the derivatives are:

• First derivative: $f'(x) = 2x \cos(x^2)$
• Second derivative: $f''(x) = 2\cos(x^2) - 4x^2 \sin(x^2)$

Example 3: Differentiating a Logarithmic Function

Differentiate $f(x) = \ln(x^2 + 1)$ and find the second derivative.

Solution:

First, find the first derivative using the chain rule:

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

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

Now, find the second derivative:

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

Using the quotient rule:

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

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

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

Thus, the derivatives are:

• First derivative: $f'(x) = \dfrac{2x}{x^2 + 1}$
• Second derivative: $f''(x) = \dfrac{2 - 2x^2}{(x^2 + 1)^2}$

Example 4: Differentiating an Exponential Function

Differentiate $f(x) = e^{3x^2 + 2x}$ and find the first and second derivatives.

Solution:

First, find the first derivative using the chain rule:

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

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

Now, find the second derivative:

$f''(x) = \dfrac{d}{dx} \left( e^{3x^2 + 2x} (6x + 2) \right)$

Using the product rule:

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

$f''(x) = e^{3x^2 + 2x} \left( (6x + 2)^2 + 6 \right)$

Thus, the derivatives are:

• First derivative: $f'(x) = e^{3x^2 + 2x} (6x + 2)$
• Second derivative: $f''(x) = e^{3x^2 + 2x} \left( (6x + 2)^2 + 6 \right)$