Higher Derivatives and Their Applications
Higher Derivatives and Their Applications

Higher-order Derivatives

## Characteristics of $$f, f’, f”$$

Given a differentiable function $$f(x)$$, we can have $$f'(x)$$ and possibly $$f”(x)$$. Each of these functions has their own characteristics, and each describes the function(s) before it.

## $$f(x)$$

$$f(x)$$ is the function in question. It may be a function of any degree and may increase and decrease at some points. The question arises when we want to know when it decreases or increases. This is the characteristic of $$f'(x)$$

## $$f'(x)$$

$$f'(x)$$ tells us the gradient of the tangent at any point of the graph, and it also tells us if the graph is increasing or decreasing at any point, including where the graph changes direction and increases after decreasing or vice versa. We call that point a critical point.

To find the tangent…

Find the equation of the tangent on graph $$y=x^2,$$ where $$x=1$$.

First, we find the gradient of the tangent:

$(x^2)’ = 2 \times x^{2-1} = 2x = 2 \times 1 = 2.$

Next, we find the value of $$y$$ when $$x=1:$$

$1^2 = 1.$

Lastly, we find which line goes through $$(1,1)$$ and has a gradient of $$2.$$ Let the equation of the gradient be $$g(x):$$

\begin{align} g(x)&=2x+b\\ 2 \times 1 + b &= 1\\ b = 1-2&=-1. \end{align}

Therefore, the tangent has equation $$2x-1.\ _\square$$

To find if the graph is increasing or decreasing at any point, we just substitute the value of $$x$$ into $$f'(x)$$. If $$f'(x)>0, f(x)$$ is increasing at $$x$$. If $$f'(x)<0, f(x)$$ is decreasing at $$x$$. If $$f'(x)=0, \big(x, f(x)\big)$$ is a critical point.

Find the critical points of $$x^3+2x^2+x+5.$$

Firstly, we find $$f'(x):$$

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

Next, we find the points where $$f'(x) = 0:$$

\begin{align} f'(x) &= (3x+1)(x+1) \\ &= 0\\ x&=-\frac{1}{3} , -1. \end{align}

Lastly, substitute the values of $$x$$ into $$f(x).$$ We get two points, $$(-1, 5)$$ and $$\big(-\frac{1}{3}, \frac{131}{27}\big).$$ $$_\square$$

## $$f”(x)$$

$$f”(x)$$ determines the concavity of the graph. A picture on it will be uploaded soon…

When $$f”(x)=0$$, concavity may change. This point(s) is(are) called inflection point(s).

With the knowledge of $$f'(x)$$ and $$f”(x)$$, graph sketching will be easier.

## Second-order Derivatives

While the tangent line is a very useful tool, when it comes to investigating the graph of a function, the tangent line fails to say anything about how the graph “bends” at a point. This is where the second derivative comes into play.

A second-order derivative is a measure of how the rate of change of a quantity is itself changing, which is obtained by differentiating again the first derivative of a quantity.

Notation:

• The second derivative of a function $$f\left( x \right)$$ is written as $$f”(x)$$.
• In Leibniz’s notation, the second derivative of $$y$$ (dependent variable) with respect to $$x$$ (independent variable) is written as $$\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }=\frac { d }{ dx } \left(\frac { dy }{ dx } \right).$$

Find the second derivative of the function $$f(x)= ax$$, where $$a$$ is any constant.

We have

$\begin{array} &f(x)=ax, &f'(x)=a, &f”(x)=0. \end{array}$

Therefore, the answer is 0. $$_\square$$

Find the second derivative of the function $$f(x)=\frac{1}{x}$$.

We have

$\begin{array} &f(x)=\dfrac{1}{x}, &f'(x)=\dfrac{-1}{x^2}, &f”(x)=\dfrac{2}{x^3}. \end{array}$

Therefore, answer is $$\frac{2}{x^3}$$. $$_\square$$

Relation with Graphs:

First, we will list all theorems and points related to graphs and second derivatives and then move to examples (from basic to advanced).

Concavity Theorem:

If the function $$f$$ is twice differentiable at $$x=a$$, then the graph of $$f$$ is concave up at $$\big(a,f(a)\big)$$ if $$f”(a)>0$$ and concave down if $$f”(a)<0$$.

Inflection Point:

An inflection point is a point on a curve where the concavity (sign) of curvature changes. Inflection points may be stationary, but they are not local maxima or minima.

To find a point of inflection, let us see an example:

(Basic)

Let $$f$$ be a function defined as $$f(x)=x^3-3x^2+2x+1$$. Find the concavity of different parts of the graph of $$f(x)$$ for different values of $$x$$. Also calculate the points of inflection.

We have $$f”(x)=6x-6$$.

For $$x>1$$, $$f”(x)>0$$. Hence the curve is concave up for $$x>1$$.

For $$x<1$$, $$f”(x)<0$$. Hence the graph is concave down for $$x<1$$.
Since the concavity is opposite for $$x>1$$ and $$x<1$$, the concavity changes at the point where $$x=1$$. $$_\square$$

Note: We can deduce that point of inflection can be obtained from $$f”(x)=0$$, which is a necessary condition for point of inflection.

(Intermediate)

Draw the graph of $$f(x)=\tan x$$ for $$x\in\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$$.

We have

$\begin{array} &f'(x)=\sec^2 x, &f”(x)=2\tan x \sec^2 x. \end{array}$

Inflection point:
Since the inflection point is obtained by setting $$f”(x)=0$$, we have $$x=0$$ because $$x\in\left(\frac{-\pi}{2},\frac{\pi}{2}\right).$$ Hence the inflection point is $$(0,0)$$.Concavity:
Since $$f”(x)=2\tan x \sec^2 x$$, we have the following:For $$x\in\left(0,\frac{\pi}{2}\right)$$, $$f”(x)>0$$. Hence it is concave up.For $$x\in\left(\frac{-\pi}{2},0\right)$$, $$f”(x)<0$$. Hence it is concave down.Result:
We have the following graph:tangent

Given $$f(x)=\tan ^{ -1 }{ \frac { 2x }{ 1-{ x }^{ 2 } } }$$, plot the graph $$y=f(x)$$.

Since

$\frac{d}{dx}\big(\tan^{-1} u\big) = \frac{1}{1+u^2},$

we can differentiate the given equation as follows:

\begin{align} \frac{dy}{dx} = \frac{d}{dx} \left( \tan ^{ -1 } \frac {2x}{1-x^2} \right) &= \frac{1}{1+\left(\frac {2x}{1-x^2} \right)^2} \cdot \frac{d}{dx} \left( \frac {2x}{1-x^2}\right) \\ &= \frac{1}{1+\frac{4x^2}{(1-x^2)^2}} \cdot \frac{2(1-x^2)-2x(-2x)}{(1-x^2)^2} \\ &= \frac{2-2x^2+4x^2}{(1-x^2)^2+4x^2} \\ &=\frac{2(1+x^2)}{(1+x^2)^2} \\ &= \frac{2}{1+x^2}. \end{align}

Again differentiating it, we get

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

Concavity:

For $$x>0$$, $$\frac{d^2y}{dx^2}=\frac { -4x }{ (1+{ x }^{ 2 })^{ 2 } }<0$$. Hence concave down.For $$x<0$$, $$\frac{d^2y}{dx^2}=\frac { -4x }{ (1+{ x }^{ 2 })^{ 2 } }>0$$. Hence concave up.

Inflection point: Setting $$\frac{d^2y}{dx^2}=\frac { -4x }{ (1+{ x }^{ 2 })^{ 2 } }=0$$ implies that the point where $$x=0$$ is the inflection point.

Result:

We get the following graph:arctan

Second Derivative Test:

A second derivative test is a test used to determine whether a stationary point where $$f'(x)=0$$ is a local maximum or minimum.

Performing the test:

$\text{If } f”(x) \begin{cases} > 0, \text{ then } f \text{ has local minima at } x. \\ = 0, \text{ then it is a possible inflection point.} \\ < 0, \text{ then } f \text{ has local maxima at } x. \end{cases}$

## Third- and Higher-order Derivatives

In order to calculate higer-order derivatives, we can use Leibniz’s formula:

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