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Second Order Derivative Calculating Higher Order Derivatives


Higher Order Derivatives



Second Order Derivative

The process of differentiation can be applied repeatedly to a function, provided the resulting derivative is itself differentiable. The derivative of the first derivative is called the second derivative. It provides information about the rate of change of the rate of change of the original function.

Definition of the Second Order Derivative

Let $y = f(x)$ be a function that is differentiable. The first derivative of $f(x)$, denoted by $f'(x)$ or $\frac{dy}{dx}$, is also a function of $x$. If this first derivative function $f'(x)$ is itself differentiable with respect to $x$, then the derivative of $f'(x)$ is called the second derivative of the original function $f(x)$.

In simple terms, the second derivative is the derivative of the derivative.


Notation for the Second Order Derivative

There are several standard notations used to represent the second derivative of $y = f(x)$ with respect to $x$:


Calculation of the Second Order Derivative

The process of finding the second derivative is straightforward:

  1. Find the first derivative. Calculate the derivative of the original function $f(x)$ using the standard differentiation rules (Power Rule, Product Rule, Quotient Rule, Chain Rule, etc.). Let the result be $f'(x)$.
  2. Find the derivative of the first derivative. Differentiate the function $f'(x)$ with respect to $x$ using the standard differentiation rules. The result of this second differentiation is the second derivative, $f''(x)$.

Essentially, you perform the differentiation process twice in succession.


Interpretation of the Second Order Derivative

The second derivative provides valuable information about the behavior of the original function and its rate of change.


Example 1. Find the second derivative of $f(x) = x^4 - 5x^2 + 7x - 1$.

Answer:

We need to find the first derivative, $f'(x)$, and then differentiate $f'(x)$ to find the second derivative, $f''(x)$.

Step 1: Find the first derivative $f'(x)$.

Differentiate the given function $f(x) = x^4 - 5x^2 + 7x - 1$ with respect to $x$ using the Power Rule, Constant Multiple Rule, Sum Rule, and Difference Rule.

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

$= \frac{d}{dx}(x^4) - \frac{d}{dx}(5x^2) + \frac{d}{dx}(7x) - \frac{d}{dx}(1)$

$= 4x^{4-1} - 5(2x^{2-1}) + 7(1x^{1-1}) - 0$

$= 4x^3 - 10x + 7$

[First derivative]

So, $f'(x) = 4x^3 - 10x + 7$.

Step 2: Find the second derivative $f''(x)$.

Differentiate the first derivative $f'(x) = 4x^3 - 10x + 7$ with respect to $x$ using the same rules.

"$f''(x) = \frac{d}{dx}(4x^3 - 10x + 7)$"

$= \frac{d}{dx}(4x^3) - \frac{d}{dx}(10x) + \frac{d}{dx}(7)$

$= 4(3x^{3-1}) - 10(1x^{1-1}) + 0$

$= 4(3x^2) - 10(1) + 0$

$= 12x^2 - 10$

[Second derivative]

The second derivative of $f(x) = x^4 - 5x^2 + 7x - 1$ is $f''(x) = 12x^2 - 10$.



Calculating Higher Order Derivatives

The derivative of a function is itself a function, provided it exists. If this derivative function is also differentiable, we can differentiate it again to obtain the second derivative. This process can be continued to find third, fourth, and even higher order derivatives, as long as the resulting functions remain differentiable.

Definition of Higher Order Derivatives

Let $y = f(x)$ be a function.

The term "higher order derivatives" usually refers to derivatives of order two or greater.


Notation for Higher Order Derivatives

Different notations are used to denote derivatives of order higher than one. The prime notation becomes cumbersome for orders higher than three, so a different convention is adopted.

The Leibniz notation $\frac{d^ny}{dx^n}$ explicitly shows the order of differentiation ($n$ in the numerator) and the variable with respect to which differentiation is done ($n$ with the $d$ in the denominator).


Process of Calculating Higher Order Derivatives

Calculating higher order derivatives is an iterative process. To find the $n$th derivative of a function $f(x)$, you simply differentiate the function $n$ times in succession with respect to the variable $x$.

  1. Start with the original function $f(x)$.
  2. Calculate the first derivative, $f'(x) = \frac{d}{dx}(f(x))$.
  3. Calculate the second derivative, $f''(x) = \frac{d}{dx}(f'(x))$.
  4. Calculate the third derivative, $f'''(x) = \frac{d}{dx}(f''(x))$.
  5. Continue this process, differentiating the previous derivative at each step, until you reach the desired $n$th derivative, $f^{(n)}(x) = \frac{d}{dx}(f^{(n-1)}(x))$.

This process relies on being able to differentiate the resulting function at each step using the standard differentiation rules (Power Rule, Product Rule, Quotient Rule, Chain Rule, etc.).


Interpretation of Higher Order Derivatives

Higher order derivatives provide increasingly detailed information about the shape and behavior of the original function's graph and the physical quantities they might represent.

Interpretation of the Second Derivative ($f''(x)$):

Interpretation of Third and Higher Derivatives:


Example 1. Find the first four derivatives of $y = \sin x$.

Answer:

We differentiate the function successively, starting with $y = \sin x$.

First Derivative ($y'$ or $\frac{dy}{dx}$):

"$y' = \frac{d}{dx}(\sin x) = \cos x$"

Second Derivative ($y''$ or $\frac{d^2y}{dx^2}$):

Differentiate the first derivative, $y' = \cos x$:

"$y'' = \frac{d}{dx}(y') = \frac{d}{dx}(\cos x) = -\sin x$"

Third Derivative ($y'''$ or $\frac{d^3y}{dx^3}$):

Differentiate the second derivative, $y'' = -\sin x$:

"$y''' = \frac{d}{dx}(y'') = \frac{d}{dx}(-\sin x) = - \frac{d}{dx}(\sin x) = -(\cos x) = -\cos x$"

Fourth Derivative ($y^{(4)}$ or $\frac{d^4y}{dx^4}$):

Differentiate the third derivative, $y''' = -\cos x$:

"$y^{(4)} = \frac{d}{dx}(y''') = \frac{d}{dx}(-\cos x) = - \frac{d}{dx}(\cos x) = -(-\sin x) = \sin x$"

The first four derivatives of $y = \sin x$ are:

  • First Derivative: $y' = \cos x$
  • Second Derivative: $y'' = -\sin x$
  • Third Derivative: $y''' = -\cos x$
  • Fourth Derivative: $y^{(4)} = \sin x$

Note that the derivatives of $\sin x$ (and similarly $\cos x$) follow a cycle of length 4: $\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x$. The 5th derivative will be the same as the 1st, the 6th the same as the 2nd, and so on.


Example 2. Find the third derivative of $f(x) = e^{2x}$.

Answer:

We need to find the first, second, and then the third derivative of $f(x) = e^{2x}$.

First Derivative ($f'(x)$):

Using the Chain Rule $\frac{d}{dx}(e^{g(x)}) = e^{g(x)} \cdot g'(x)$ with $g(x)=2x$ and $g'(x)=2$:

"$f'(x) = \frac{d}{dx}(e^{2x}) = e^{2x} \cdot \frac{d}{dx}(2x) = e^{2x} \cdot 2 = 2e^{2x}$"

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

Second Derivative ($f''(x)$):

Differentiate $f'(x) = 2e^{2x}$. Use the Constant Multiple Rule and the Chain Rule again:

"$f''(x) = \frac{d}{dx}(2e^{2x}) = 2 \cdot \frac{d}{dx}(e^{2x})$"

From the previous step, we know $\frac{d}{dx}(e^{2x}) = 2e^{2x}$:

$= 2 \cdot (2e^{2x}) = 4e^{2x}$"

So, $f''(x) = 4e^{2x}$.

Third Derivative ($f'''(x)$):

Differentiate $f''(x) = 4e^{2x}$. Use the Constant Multiple Rule and the Chain Rule:

"$f'''(x) = \frac{d}{dx}(4e^{2x}) = 4 \cdot \frac{d}{dx}(e^{2x})$"

Again, using $\frac{d}{dx}(e^{2x}) = 2e^{2x}$:

$= 4 \cdot (2e^{2x}) = 8e^{2x}$"

The third derivative of $f(x) = e^{2x}$ is $f'''(x) = 8e^{2x}$.

Observe the pattern: $f^{(n)}(x) = 2^n e^{2x}$.


Example 3. If $y = x^5 + 2x^4 - 6x^3 + x^2 - 9x + 15$, find $\frac{d^6y}{dx^6}$.

Answer:

We need to find the sixth derivative of the given polynomial function by differentiating six times in succession.

First Derivative ($\frac{dy}{dx}$):

"$\frac{dy}{dx} = \frac{d}{dx}(x^5 + 2x^4 - 6x^3 + x^2 - 9x + 15) = 5x^4 + 8x^3 - 18x^2 + 2x - 9$"

Second Derivative ($\frac{d^2y}{dx^2}$):

"$\frac{d^2y}{dx^2} = \frac{d}{dx}(5x^4 + 8x^3 - 18x^2 + 2x - 9) = 20x^3 + 24x^2 - 36x + 2$"

Third Derivative ($\frac{d^3y}{dx^3}$):

"$\frac{d^3y}{dx^3} = \frac{d}{dx}(20x^3 + 24x^2 - 36x + 2) = 60x^2 + 48x - 36$"

Fourth Derivative ($\frac{d^4y}{dx^4}$):

"$\frac{d^4y}{dx^4} = \frac{d}{dx}(60x^2 + 48x - 36) = 120x + 48$"

Fifth Derivative ($\frac{d^5y}{dx^5}$):

"$\frac{d^5y}{dx^5} = \frac{d}{dx}(120x + 48) = 120$"

Sixth Derivative ($\frac{d^6y}{dx^6}$):

The fifth derivative is a constant, 120. The derivative of any constant is 0.

"$\frac{d^6y}{dx^6} = \frac{d}{dx}(120) = 0$"

[Derivative of a constant is 0]

The sixth derivative is 0.

In general, for any polynomial of degree $n$, the $(n+1)$th derivative and all subsequent higher order derivatives are always zero. In this case, the polynomial has degree 5, so the 6th derivative is 0.