Third Derivative Calculator

To find the third derivative of the function sin⁡(2x), you need to apply the rules of differentiation. Let's find its solution in detail. The given function is,

$$ f(x) \;=\; sin (2x) $$

Use chain rule for differentiation, then

$$ \frac{d}{dx} g(h(x)) \;=\; g’(h(x)) . h’(x) $$

Differentiate the given function f(x) with respect to x.

$$ f’(x) \;=\; \frac{d}{dx} sin(2x) \;=\; cos(2x) \;=\; cos(2x) . 2 \;=\; 2 cos(2x) $$

Again differentiate the f`(x) with respect to x,

$$ f’’(x) \;=\; \frac{d}{dx} [2 cos(2x)] \;=\; 2 . \frac{d}{dx} cos(2x) \;=\; 2 . (-sin(2x)) . \frac{d}{dx} (2x) \;=\; 2 . (-sin(2x)) . 2 $$

For the solution of the third derivation, again take the derivative of f``(x) second order derivative function,

$$ f’’’(x) \;=\; \frac{d}{dx} -4 sin(2x) = -4 . \frac{d}{dx} sin(2x) = -4 . cos(2x) . \frac{d}{dx} (2x) \;=\; -4 . cos(2x) . 2 $$

Hence the solution of the given third derivative function is,

$$ f’’’(x) \;=\; -8 cos(2x) $$

If the first, second, and third derivatives of a function f(x) are all zero, it expresses a specific situation about the nature of the function f(x). Let's analyze this in steps.

• First Derivative f′(x) = 0:

The function f(x) is a constant function because the slope of the function is zero everywhere. So, f(x) does not change as x changes.

• Second Derivative f′′(x) = 0:

Since f′(x) = 0, f'(x) = 0, f′′(x) = 0 says that f′(x) is not only zero at specific points but is zero everywhere so that f(x) is a constant function.

• Third Derivative f′′′(x) = 0

The third derivative being zero is consistent as the first and second derivatives is zero everywhere, further, it confirms that f(x) is a constant function.

Conclusion

If f′(x) = 0, f′′(x) = 0, and f′′′(x) = 0 for all x, then the function f(x) must be a constant function. In other words, f(x) = C where C is a constant.

The third derivative of a function has important purposes in both theoretical and applied mathematics. It provides insights into the behavior and properties of the function. Here are some of the key purposes and applications of the third derivative:

• Rate of Change of Curvature:

The third derivative f′′′(x) measures the rate at which this curvature is changing. to understand the nature of points of inflection and the dynamic behavior of curves.

• Jerk in Kinematics:

In physics, the third derivative is used to study the motion of the position function with respect to time and is known as "jerk" or "jolt". It represents the rate of change of acceleration.

• Taylor Series Expansion:

The third derivative plays a role in the Taylor series expansion of a function.

• Study of Oscillatory Behavior:

In the analysis of oscillatory systems, such as electrical circuits or mechanical vibrations, the third derivative can provide insights into the behavior of the system.

• Smoothing and Filtering:

In signal processing, higher-order derivatives, including the third derivative, are used in the design of filters and in the analysis of signals.

• Modeling and Analysis of Financial Data:

In finance, the third derivative can be used in the modeling and analysis of financial data, such as in the study of the sensitivity of option prices.

To find the third derivative of the function f(x) = 1/x, differentiate it three times. For differentiation use the power rule with respect to x,

$$ \frac{1}{x} \;=\; x^{-1} $$

$$ f(x) \;=\; x^{-1} $$

$$ f’(x) \;=\; \frac{d}{dx}(x^{-1}) = -x^{-2} $$

$$ f’(x) \;=\; - \frac{1}{x^2} $$

Again differentiate f`(x) with respect to x,

$$ f’(x) \;=\; -x^2 $$

$$ f’’(x) \;=\; \frac{d}{dx} (-x^2) \;=\; -(-2)x^{-3} $$

$$ f’’(x) \;=\; 2x^{-3} $$

$$ f’’(x) \;=\; \frac{2}{x^3} $$

To get the solution of the third derivative solution, differentiate the second derivative function f``(x) with respect to x.

$$ f’’(x) \;=\; 2x^{-3} $$

$$ f’’’(x) \;=\; \frac{d}{dx}(2x^{-3}) \;=\; 2 . (-3)x^{-4} $$

$$ f’’’(x) \;=\; - 6x^{-4} $$

$$ f’’’(x) \;=\; - \frac{6}{x^4} $$

Therefore the solution of the given third derivative function 1/x is,

$$ f’’’(x) \;=\; - \frac{6}{x^4} $$

To find the third derivative of the function f(x) = tan⁻¹(x), differentiate it by using the appropriate function three times.

Differentiate f(x) with respect to x.

The given function tan^-1x has the direct formula of differentiation,

$$ f’(x) \;=\; \frac{d}{dx} tan^{-1} (x) \;=\; \frac{1}{1+x^2} $$

Again differentiate f`(x) with respect to x,

$$ f’(x) \;=\; \frac{1}{1+x^2} $$

Use the chain rule or power rule for differentiation,

$$ f’(x) \;=\; (1 + x^2)^{-1} $$

$$ f’’(x) \;=\; \frac{d}{dx} (1 + x^2)^{-1} \;=\; -1 . (1 + x^2)^{-2} . \frac{d}{dx} (1 + x^2) $$

$$ f’’(x) \;=\; -(1 + x^2)^{-2} . 2x $$

$$ f’’(x) \;=\; - \frac{2x}{(1+x^2)^2} $$

Again differentiate f``(x) with respect to x to get the solution of the third derivative function,

$$ f’’(x) \;=\; -2x(1 + x^2)^{-2} $$

Use the product rule or chain rule for derivation,

$$ f’’’(x) \;=\; -2\left[(1 + x^2)^{-2} + x . \frac{d}{dx}(1 + x^2)^{-2} \right] $$

$$ f’’’(x) \;=\; -2 \left[(1 + x^2)^{-2} + x . (-4x(1 + x^2)^{-3}) \right] $$

$$ f’’’(x) \;=\; -2[(1 + x^2)^{-2} - 4x^2(1 + x^2)^{-3}] $$

Factor out (1 + x²)^-3:

$$ f’’’(x) \;=\; -2 [(1 + x^2)^{-2} (1 + x^2) . (1 + x^2)^{-1} - 4x^2(1 + x^2)^{-3}] $$

$$ f’’’(x) \;=\; -2 [(1 + x^2 - 4x^2)(1 + x^2)^{-3}] $$

$$ f’’’(x) \;=\; -2[(1 - 3x^2)(1 + x^2)^{-3}] $$

$$ f’’’(x) \;=\; -2 \frac{1 - 3x^2}{(1 + x^2)^3} $$

Therefore the solution of the given third derivative function tan^-1x is,

$$ f’’’(x) \;=\; -2 \frac{1 - 3x^2}{(1 + x^2)^3} $$

To find the product rule of differentiation, feel free to use the product rule solver.