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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 $x$:
- Prime Notation:
- $f''(x)$ (read as "f double prime of x" or "f double dash of x")
- $y''$ (read as "y double prime" or "y double dash", if $y=f(x)$)
- Leibniz Notation:
- $\frac{d^2y}{dx^2}$ (read as "d squared y by dx squared" or "the second derivative of y with respect to x")
- $\frac{d^2}{dx^2} f(x)$ (read as "d squared by dx squared of f of x")
This notation emphasizes that the differentiation operator $\frac{d}{dx}$ is applied twice to the function $y$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$
- Operator Notation:
- $D^2(f(x))$ or $D^2y$ (where $D = \frac{d}{dx}$)
Calculation of the Second Order Derivative
The process of finding the second derivative is straightforward:
- 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 $f'(x)$.
- 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.
- Rate of Change of the Rate of Change: The second derivative $f''(x)$ tells us how the slope of the tangent line (given by $f'(x)$) is changing as $x$ changes.
- If $f''(x) > 0$, the slope $f'(x)$ is increasing.
- If $f''(x) < 0$, the slope $f'(x)$ is decreasing.
- Concavity of the Graph: The sign of the second derivative determines the concavity of the graph of $y=f(x)$:
- If $f''(x) > 0$ on an interval, the graph of $f(x)$ is concave upward (or simply concave up) on that interval. The curve bends upwards, like a cup that can hold water.
- If $f''(x) < 0$ on an interval, the graph of $f(x)$ is concave downward (or simply concave down) on that interval. The curve bends downwards, like a cup that spills water.
- Physics (Motion Analysis): In the context of motion, if $s(t)$ represents the position of an object along a line at time $t$:
- The first derivative $s'(t) = v(t)$ is the instantaneous velocity.
- The second derivative $s''(t) = v'(t) = a(t)$ is the instantaneous acceleration. Acceleration is the rate of change of velocity. A positive acceleration means velocity is increasing, while a negative acceleration means velocity is decreasing (deceleration).
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 first derivative is $f'(x) = \frac{dy}{dx}$.
- The second derivative is the derivative of the first derivative: $f''(x) = \frac{d}{dx}(f'(x))$.
- The third derivative is the derivative of the second derivative: $f'''(x) = \frac{d}{dx}(f''(x))$.
- The fourth derivative is the derivative of the third derivative: $f^{(4)}(x) = \frac{d}{dx}(f'''(x))$.
- In general, the $n$th derivative (for $n \ge 2$) is the derivative of the $(n-1)$th derivative: $f^{(n)}(x) = \frac{d}{dx}(f^{(n-1)}(x))$, provided $f^{(n-1)}(x)$ is differentiable.
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.
- Second Derivative:
$f''(x)$, $y''$, $\frac{d^2y}{dx^2}$, $D^2y$
- Third Derivative:
$f'''(x)$, $y'''$, $\frac{d^3y}{dx^3}$, $D^3y$
- Fourth Derivative: (Parentheses indicate the order, not a power)
$f^{(4)}(x)$, $y^{(4)}$, $\frac{d^4y}{dx^4}$, $D^4y$
- Fifth Derivative:
$f^{(5)}(x)$, $y^{(5)}$, $\frac{d^5y}{dx^5}$, $D^5y$
- $n$th Derivative: (For $n \ge 4$)
$f^{(n)}(x)$, $y^{(n)}$, $\frac{d^ny}{dx^n}$, $D^ny$
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 $x$.
- Start with the original function $f(x)$.
- Calculate the first derivative, $f'(x) = \frac{d}{dx}(f(x))$.
- Calculate the second derivative, $f''(x) = \frac{d}{dx}(f'(x))$.
- Calculate the third derivative, $f'''(x) = \frac{d}{dx}(f''(x))$.
- 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 ($f''(x)$):
- Rate of Change of Slope: $f''(x)$ represents how the slope of the tangent line, $f'(x)$, is changing as $x$ varies. If $f''(x)$ is large and positive, the slope is increasing rapidly. If $f''(x)$ is large and negative, the slope is decreasing rapidly.
- Concavity: This is a primary geometric interpretation. The sign of the second derivative indicates the bending direction of the curve:
- If $f''(x) > 0$ on an interval, the graph of $f(x)$ is concave upward on that interval. The curve "holds water".
- If $f''(x) < 0$ on an interval, the graph of $f(x)$ is concave downward on that interval. The curve "spills water".
- Acceleration in Physics: If $s(t)$ is the position of an object at time $t$, then $s'(t) = v(t)$ is velocity, and $s''(t) = v'(t) = a(t)$ is the acceleration. Acceleration is the rate at which velocity changes. Positive acceleration means velocity is increasing (speeding up in the positive direction or slowing down in the negative direction), while negative acceleration means velocity is decreasing (slowing down in the positive direction or speeding up in the negative direction).
Interpretation of Third and Higher Derivatives:
- Third Derivative ($f'''(x)$): Represents the rate of change of acceleration (sometimes called "jerk" in physics). Geometrically, it relates to how the concavity is changing. Points where the concavity changes ($f''(x)$ changes sign) are often inflection points, and the third derivative can help confirm these points (if $f''(c)=0$ and $f'''(c) \neq 0$, then $c$ is an inflection point).
- Higher Order Derivatives: These are used in advanced calculus topics such as Taylor and Maclaurin series expansions, which approximate functions using polynomials derived from the function's derivatives at a single point. They also appear in the study and solution of higher-order differential equations, which model more complex systems in science and engineering.
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.