If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka

# Differential Calculus

I greet you this day,
Second: view the videos.
Third: solve the solved examples and word problems
Fourth: check your solutions with my thoroughly-explained solutions.
Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome.
If you are my student, please do not contact me here. Contact me via the school's system. Thank you for visiting.

Samuel Dominic Chukwuemeka (Samdom For Peace) B.Eng., A.A.T, M.Ed., M.S

## Objectives

Students will:

(1.) Discuss the concept of the derivative of a function.

(2.) Determine the derivative of an explicit function by limit.

(3.) Determine the derivative of an explicit function by rules.

(4.) Determine the higher-order derivatives of explicit functions.

(5.) Differentiate implicit functions.

(6.) Differentiate exponential functions.

(7.) Differentiate logarithmic functions.

(8.) Differentiate trigonometric functions.

(9.) Solve applied problems involving the derivatives of functions.

## Power Rule

Pre-requisite Topic: Exponents

The Power Rule of Derivatives states that:

$If\:\: y = x^n \\[3ex] then\:\: \dfrac{dy}{dx} = nx^{n - 1} \\[5ex] If\:\: y = ax^n \\[3ex] then\:\: \dfrac{dy}{dx} = nax^{n - 1}$

## Sum/Difference Rule

The Sum Rule of Derivatives states that:

$If\:\: y = u + v \\[3ex] where\:\: u = f(x);\:\: v = f(x) \\[3ex] then\:\: \dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dv}{dx} \\[5ex] If\:\: y = u + v + w \\[3ex] where\:\: u = f(x);\:\: v = f(x);\:\: w = f(x) \\[3ex] then\:\: \dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dv}{dx} + \dfrac{dw}{dx}$

The Difference Rule of Derivatives states that:

$If\:\: y = u - v \\[3ex] where\:\: u = f(x);\:\: v = f(x) \\[3ex] then\:\: \dfrac{dy}{dx} = \dfrac{du}{dx} - \dfrac{dv}{dx} \\[5ex] If\:\: y = u - v - w \\[3ex] where\:\: u = f(x);\:\: v = f(x);\:\: w = f(x) \\[3ex] then\:\: \dfrac{dy}{dx} = \dfrac{du}{dx} - \dfrac{dv}{dx} - \dfrac{dw}{dx}$

You can write it in a compact form
The Sum/Difference Rule of Derivatives states that:

$If\:\: y = u \pm v \\[3ex] where\:\: u = f(x);\:\: v = f(x) \\[3ex] then\:\: \dfrac{dy}{dx} = \dfrac{du}{dx} \pm \dfrac{dv}{dx} \\[5ex] If\:\: y = u \pm v \pm w \\[3ex] where\:\: u = f(x);\:\: v = f(x);\:\: w = f(x) \\[3ex] then\:\: \dfrac{dy}{dx} = \dfrac{du}{dx} \pm \dfrac{dv}{dx} \pm \dfrac{dw}{dx}$

## Chain Rule

The Chain Rule is also known as the Function of a Function Rule
We shall see the reasons.

Let us begin with the reason why it is called the Function of a Function Rule
$y$ is not a direct function of $x$
Rather, it is a function of some other function that is a function of $x$

The Function of a Function Rule of Derivatives states that:

$If\:\: y = f(u) \\[3ex] and\:\: u = f(x) \\[3ex] then\:\: \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} \\[5ex]$ As you can see, $y$ is a function of $u$ which is then a function of $x$
So, $y$ is a function of a function of $x$

Student: Does this apply only to the case of three variables - $y, u, x$
where $y$ is a function of $u$ which is a function of $x$
Teacher: Good question.
No, it does not apply to only three variables.
It can apply to so many variables - sort of like a chain
Hence, the name "Chain Rule"

The Chain Rule of Derivatives states that:

$If\:\: y = f(u) \\[3ex] and\:\: u = f(v) \\[3ex] and\:\: v = f(x) \\[3ex] then\:\: \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dv} * \dfrac{dv}{dx} \\[5ex]$ As you can see, this is a case of $3$ chains

$If\:\: y = f(u) \\[3ex] and\:\: u = f(v) \\[3ex] and\:\: v = f(w) \\[3ex] and\:\: w = f(x) \\[3ex] then\:\: \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dv} * \dfrac{dv}{dw} * \dfrac{dw}{dx} \\[5ex]$ This is a case of $4$ chains

Student: How do we set up the chains?
How do we know we set it up correctly?
How do we know when to use it?
Teacher: Good question.
When you see composite functions or functions with several exponents or functions that are not "direct" sums, differences, product, or quotient; then the Chain Rule is probably the likely rule to use
You do want to break it up into several functions as simply as possible
After setting it up and multiplying the functions, it will result to $\dfrac{dy}{dx}$

All the terms will cancel out leaving only the $\dfrac{dy}{dx}$

That way, you know you set it up correctly.
If you review the $2$ chains, $3$ chains, and $4$ chains, you will notice that only the $\dfrac{dy}{dx}$ remains.

## Product Rule

The Product Rule of Derivatives states that:

$If\:\: y = u * v \\[3ex] where\:\: u = f(x);\:\: v = f(x) \\[3ex] then\:\: \dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx} \\[5ex]$ NOTE: Before using the Product Rule, you must first express the function as the product of only two functions.
In other words, it must be the product of only two functions.

Pnuemonic to Remember Product Rule
Let: $y = u * v \\[3ex] u = first \\[3ex] v = second \\[3ex] y = first * second \\[3ex]$ $\dfrac{dy}{dx} = first * dee-second + second * dee-first$

## Quotient Rule

The Quotient Rule of Derivatives states that:

$If\:\: y = \dfrac{u}{v} \\[5ex] where\:\: u = f(x);\:\: v = f(x) \\[3ex] then\:\: \dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2} \\[5ex]$ NOTE: Before using the Quotient Rule, you must first express the function as the quotient of only two functions.
In other words, it must be the quotient of only two functions.

Pnuemonic to Remember Quotient Rule
Let: $y = \dfrac{u}{v} \\[5ex] u = top \\[3ex] v = bottom \\[3ex] y = \dfrac{top}{bottom} \\[5ex]$ $\dfrac{dy}{dx} = [(bottom * dee-top) \:\:minus\:\: (top * dee-bottom)] \:\:all\:\:over\:\: bottom-squared$

## Derivatives of Basic Trigonometric Functions

Pre-requisite Knowledge (from the Pre-requisite Topics):

$(1.)\:\: \displaystyle{\lim_{\theta \to 0}} \dfrac{\sin\theta}{\theta} = 1...Special\:\:Limit \\[5ex] (2.)\:\: \sin \alpha - \sin \beta = 2 \sin \left(\dfrac{\alpha - \beta}{2}\right) \cos \left(\dfrac{\alpha + \beta}{2}\right)...Sum-to-Product\:\:Formula \\[5ex] (3.)\:\: \cos \alpha - \cos \beta = -2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \sin \left(\dfrac{\alpha - \beta}{2}\right)...Sum-to-Product\:\:Formula \\[5ex] (4.)\:\: \tan\theta = \dfrac{\sin\theta}{\cos\theta} ...Quotient\:\:Identity \\[5ex] (5.)\:\: \sin^2 \theta + \cos^2 \theta = 1...Pythagorean\:\:Identity \\[3ex] (6.)\:\: \sec\theta = \dfrac{1}{\cos\theta}...Reciprocal\:\:Identity \\[5ex] (7.)\:\: \csc\theta = \dfrac{1}{\sin\theta}...Reciprocal\:\:Identity \\[5ex] (8.)\:\: \cot\theta = \dfrac{\cos\theta}{\sin\theta}$

The basic trigonometric functions are:

$(1.)\:\: \sin x \\[3ex] (2.)\:\: \cos x \\[3ex] (3.)\:\: \tan x \\[3ex] (4.)\:\: \csc x \\[3ex] (5.)\:\: \sec x \\[3ex] (6.)\:\: \cot x$

(1.) Derivative of $\boldsymbol{\sin x}$
We shall use the Special Limit for $\displaystyle{\lim_{\theta \to 0}} \dfrac{\sin\theta}{\theta}$, Derivatives by Limits, and Trigonometric Formulas (Sum-to-Product Formulas)
Based on Derivatives by Limits (Derivatives from First Principle)

$y = \sin x \\[3ex] y + \Delta y = \sin (x + \Delta x) \\[3ex] \Delta y = \sin (x + \Delta x) - y \\[3ex] \Delta y = \sin (x + \Delta x) - \sin x \\[3ex] \sin \alpha - \sin \beta = 2 \sin \left(\dfrac{\alpha - \beta}{2}\right) \cos \left(\dfrac{\alpha + \beta}{2}\right)...Sum-to-Product\:\:Formula \\[5ex] Let\:\: \alpha = x + \Delta x \\[3ex] Let\:\: \beta = x \\[3ex] \rightarrow \sin (x + \Delta x) - \sin x = 2 \sin \left(\dfrac{x + \Delta x - x}{2}\right) \cos \left(\dfrac{x + \Delta x + x}{2}\right)...Sum-to-Product\:\:Formula \\[5ex] \sin (x + \Delta x) - \sin x = 2 \sin \left(\dfrac{\Delta x}{2}\right) \cos \left(\dfrac{2x + \Delta x}{2}\right) \\[5ex] 2x + \Delta x = 2\left(x + \dfrac{\Delta x}{2}\right) ...Factor\:\:by\:\:2 \\[5ex] \dfrac{2x + \Delta x}{2} = \dfrac{2\left(x + \dfrac{\Delta x}{2}\right)}{2} = x + \dfrac{\Delta x}{2} \\[5ex] \rightarrow \sin (x + \Delta x) - \sin x = 2 \sin \left(\dfrac{\Delta x}{2}\right) \cos \left(x + \dfrac{\Delta x}{2}\right) \\[5ex] \rightarrow \Delta y = 2 \sin \left(\dfrac{\Delta x}{2}\right) \cos \left(x + \dfrac{\Delta x}{2}\right) \\[5ex] Divide\:\:both\:\:sides\:\:by\:\:\Delta x \\[3ex] \dfrac{\Delta y}{\Delta x} = \dfrac{2 \sin \left(\dfrac{\Delta x}{2}\right) \cos \left(x + \dfrac{\Delta x}{2}\right)}{\Delta x} \\[7ex] Multiply\:\:both\:\:the\:\:numerator\:\:and\:\:denominator\:\:on\:\:the\:\:RHS\:\:by\:\:\dfrac{1}{2} \\[5ex] \dfrac{2 * \dfrac{1}{2} * \sin \left(\dfrac{\Delta x}{2}\right) * \cos \left(x + \dfrac{\Delta x}{2}\right)}{\dfrac{1}{2} * \Delta x} = \dfrac{\sin \left(\dfrac{\Delta x}{2}\right) \cos \left(x + \dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} \\[7ex] \rightarrow \dfrac{\Delta y}{\Delta x} = \dfrac{\sin \left(\dfrac{\Delta x}{2}\right) \cos \left(x + \dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} \\[7ex] \dfrac{\Delta y}{\Delta x} = \dfrac{\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} * \cos\left(x + \dfrac{\Delta x}{2}\right) \\[7ex] Introduce\:\:limits \\[3ex] \displaystyle{\lim_{\Delta x \to 0}}\dfrac{\Delta y}{\Delta x} = \displaystyle{\lim_{\theta \to 0}}\dfrac{\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} * \displaystyle{\lim_{\theta \to 0}}\cos \left(x + \dfrac{\Delta x}{2}\right) \\[7ex] \displaystyle{\lim_{\Delta x \to 0}}\dfrac{\Delta y}{\Delta x} = \dfrac{dy}{dx} \\[5ex] \displaystyle{\lim_{\Delta x \to 0}}\dfrac{\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} = \displaystyle{\lim_{\Delta x \to 0}}\dfrac{\sin \left(\dfrac{0}{2}\right)}{\dfrac{0}{2}} = \displaystyle{\lim_{\Delta x \to 0}}\dfrac{\sin 0}{0} = 1 \\[7ex] \displaystyle{\lim_{\Delta x \to 0}}\cos\left(x + \dfrac{\Delta x}{2}\right) = \displaystyle{\lim_{\Delta x \to 0}}\cos\left(x + \dfrac{0}{2}\right) = \displaystyle{\lim_{\Delta x \to 0}}\cos(x + 0) = \cos x \\[7ex] \rightarrow \dfrac{dy}{dx} = 1 * \cos x \\[5ex] \therefore \dfrac{dy}{dx} = \cos x$

(2.) Derivative of $\boldsymbol{\cos x}$
We shall use the Special Limit for $\displaystyle{\lim_{\theta \to 0}} \dfrac{\sin\theta}{\theta}$, Derivatives by Limits, and Trigonometric Formulas (Sum-to-Product Formulas)
Based on Derivatives by Limits (Derivatives from First Principle)

$y = \cos x \\[3ex] y + \Delta y = \cos(x + \Delta x) \\[3ex] \Delta y = \cos(x + \Delta x) - y \\[3ex] \Delta y = \cos(x + \Delta x) - \cos x \\[3ex] \cos \alpha - \cos \beta = -2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \sin \left(\dfrac{\alpha - \beta}{2}\right)...Sum-to-Product\:\:Formula \\[5ex] Let\:\: \alpha = x + \Delta x \\[3ex] Let\:\: \beta = x \\[3ex] \rightarrow \cos(x + \Delta x) - \cos x = -2 \sin \left(\dfrac{x + \Delta x + x}{2}\right) \sin \left(\dfrac{x + \Delta x - x}{2}\right)...Sum-to-Product\:\:Formula \\[5ex] \cos(x + \Delta x) - \cos x = -2 \sin \left(\dfrac{2x + \Delta x}{2}\right) \sin \left(\dfrac{\Delta x}{2}\right) \\[5ex] 2x + \Delta x = 2\left(x + \dfrac{\Delta x}{2}\right) ...Factor\:\:by\:\:2 \\[5ex] \dfrac{2x + \Delta x}{2} = \dfrac{2\left(x + \dfrac{\Delta x}{2}\right)}{2} = x + \dfrac{\Delta x}{2} \\[5ex] \rightarrow \cos(x + \Delta x) - \cos x = -2\sin \left(x + \dfrac{\Delta x}{2}\right)\sin \left(\dfrac{\Delta x}{2}\right) \\[5ex] \rightarrow \Delta y = -2\sin \left(x + \dfrac{\Delta x}{2}\right)\sin \left(\dfrac{\Delta x}{2}\right) \\[5ex] Divide\:\:both\:\:sides\:\:by\:\:\Delta x \\[3ex] \dfrac{\Delta y}{\Delta x} = \dfrac{-2\sin \left(x + \dfrac{\Delta x}{2}\right)\sin \left(\dfrac{\Delta x}{2}\right)}{\Delta x} \\[7ex] Multiply\:\:both\:\:the\:\:numerator\:\:and\:\:denominator\:\:on\:\:the\:\:RHS\:\:by\:\:\dfrac{1}{2} \\[5ex] \dfrac{-2 * \dfrac{1}{2} * \sin \left(x + \dfrac{\Delta x}{2}\right)\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{1}{2} * \Delta x} = \dfrac{-\sin \left(x + \dfrac{\Delta x}{2}\right)\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} \\[7ex] \rightarrow \dfrac{\Delta y}{\Delta x} = \dfrac{-\sin \left(x + \dfrac{\Delta x}{2}\right)\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} \\[7ex] \dfrac{\Delta y}{\Delta x} = -\sin \left(x + \dfrac{\Delta x}{2}\right) * \dfrac{\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} \\[7ex] Introduce\:\:limits \\[3ex] \displaystyle{\lim_{\Delta x \to 0}}\dfrac{\Delta y}{\Delta x} = \displaystyle{\lim_{\theta \to 0}}-\sin \left(x + \dfrac{\Delta x}{2}\right) * \displaystyle{\lim_{\theta \to 0}}\dfrac{\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} \\[7ex] \displaystyle{\lim_{\Delta x \to 0}}\dfrac{\Delta y}{\Delta x} = \dfrac{dy}{dx} \\[5ex] \displaystyle{\lim_{\Delta x \to 0}}-\sin\left(x + \dfrac{\Delta x}{2}\right) = \displaystyle{\lim_{\Delta x \to 0}}-\sin\left(x + \dfrac{0}{2}\right) = \displaystyle{\lim_{\Delta x \to 0}}-\sin(x + 0) = -\sin x \\[7ex] \displaystyle{\lim_{\Delta x \to 0}}\dfrac{\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} = \displaystyle{\lim_{\Delta x \to 0}}\dfrac{\sin \left(\dfrac{0}{2}\right)}{\dfrac{0}{2}} = \displaystyle{\lim_{\Delta x \to 0}}\dfrac{\sin 0}{0} = 1 \\[7ex] \rightarrow \dfrac{dy}{dx} = -\sin x * 1 \\[5ex] \therefore \dfrac{dy}{dx} = -\sin x$

(3.) Derivative of $\boldsymbol{\tan x}$
We shall use the Trigonometric Identities and the Quotient Rule

$y = \tan x = \dfrac{\sin x}{\cos x} ...Quotient\:\:Identity \\[5ex] \dfrac{\sin x}{\cos x} = \dfrac{u}{v} \\[5ex] u = \sin x \\[3ex] \dfrac{du}{dx} = \cos x \\[5ex] v = \cos x \\[3ex] \dfrac{dv}{dx} = -\sin x \\[5ex] \dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2} ...Quotient\:\:Rule \\[5ex] \dfrac{dy}{dx} = \dfrac{\cos x(\cos x) - \sin x(-\sin x)}{\cos^2x} \\[5ex] \dfrac{dy}{dx} = \dfrac{\cos^2x + \sin^2x}{\cos^2x} \\[5ex] \cos^2x + \sin^2x = 1 ...Pythagorean\:\:Identity \\[3ex] \rightarrow \dfrac{dy}{dx} = \dfrac{1}{\cos^2x} \\[5ex] \dfrac{1}{\cos x} = \sec x ...Reciprocal\:\:Identity \\[5ex] \rightarrow \dfrac{1^2}{\cos^2x} = \dfrac{1}{\cos^2x} = \sec^2x \\[5ex] \therefore \dfrac{dy}{dx} = \sec^2x$

(4.) Derivative of $\boldsymbol{\csc x}$
We shall use the Trigonometric Identities and the Quotient Rule

$y = \csc x = \dfrac{1}{\sin x} ...Reciprocal\:\:Identity \\[5ex] \dfrac{1}{\sin x} = \dfrac{u}{v} \\[5ex] u = 1 \\[3ex] \dfrac{du}{dx} = 0 \\[5ex] v = \sin x \\[3ex] \dfrac{dv}{dx} = \cos x \\[5ex] \dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2} ...Quotient\:\:Rule \\[5ex] \dfrac{dy}{dx} = \dfrac{\sin x(0) - 1(\cos x)}{\sin^2x} \\[5ex] \dfrac{dy}{dx} = \dfrac{0 - \cos x}{\sin^2x} = \dfrac{-\cos x}{\sin^2x} = \dfrac{-\cos x}{\sin x} * \dfrac{1}{\sin x} \\[5ex] \dfrac{-\cos x}{\sin x} = -\cot x ...Quotient\:\:Identity \\[5ex] \dfrac{1}{\sin x} = \csc x ...Reciprocal\:\:Identity \\[5ex] \rightarrow \dfrac{dy}{dx} = -\cot x * \csc x \\[5ex] \therefore \dfrac{dy}{dx} = -\cot x\csc x$

(5.) Derivative of $\boldsymbol{\sec x}$
We shall use the Trigonometric Identities and the Quotient Rule

$y = \sec x = \dfrac{1}{\cos x} ...Reciprocal\:\:Identity \\[5ex] \dfrac{1}{\cos x} = \dfrac{u}{v} \\[5ex] u = 1 \\[3ex] \dfrac{du}{dx} = 0 \\[5ex] v = \cos x \\[3ex] \dfrac{dv}{dx} = -\sin x \\[5ex] \dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2} ...Quotient\:\:Rule \\[5ex] \dfrac{dy}{dx} = \dfrac{\cos x(0) - 1(-\sin x)}{\cos^2x} \\[5ex] \dfrac{dy}{dx} = \dfrac{0 + \sin x}{\cos^2x} = \dfrac{\sin x}{\cos^2x} = \dfrac{\sin x}{\cos x} * \dfrac{1}{\cos x} \\[5ex] \dfrac{\sin x}{\cos x} = \tan x ...Quotient\:\:Identity \\[5ex] \dfrac{1}{\cos x} = \sec x ...Reciprocal\:\:Identity \\[5ex] \rightarrow \dfrac{dy}{dx} = \tan x * \sec x \\[5ex] \therefore \dfrac{dy}{dx} = \tan x\sec x$

(6.) Derivative of $\boldsymbol{\cot x}$
We shall use the Trigonometric Identities and the Quotient Rule

$y = \cot x = \dfrac{\cos x}{\sin x} ...Quotient\:\:Identity \\[5ex] \dfrac{\cos x}{\sin x} = \dfrac{u}{v} \\[5ex] u = \cos x \\[3ex] \dfrac{du}{dx} = -\sin x \\[5ex] v = \sin x \\[3ex] \dfrac{dv}{dx} = \cos x \\[5ex] \dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2} ...Quotient\:\:Rule \\[5ex] \dfrac{dy}{dx} = \dfrac{\sin x(-\sin x) - \cos x(\cos x)}{\sin^2x} \\[5ex] \dfrac{dy}{dx} = \dfrac{-\sin^2x - \cos^2x}{\sin^2x} = \dfrac{-1(\sin^2x + \cos^2x)}{\sin^2x} \\[5ex] \sin^2x + \cos^2x = 1 ...Pythagorean\:\:Identity \\[3ex] \rightarrow \dfrac{dy}{dx} = \dfrac{-1}{\sin^2x} = -1 * \dfrac{1}{\sin^2x} \\[5ex] \dfrac{1}{\sin x} = \csc x ...Reciprocal\:\:Identity \\[5ex] \rightarrow \dfrac{1^2}{\sin^2x} = \dfrac{1}{\sin^2x} = \csc^2x \\[5ex] \rightarrow \dfrac{dy}{dx} = -1 * \csc^2x \\[5ex] \therefore \dfrac{dy}{dx} = -\csc^2x$

## Implicit Differentiation

Implicit Differentiation or Implicit Differential Calculus is the derivative of implicit functions.
An Implicit Function is a function where the dependent and independent variables are expressed implicitly. In other words, the dependent variable is not expressed in terms of the independent variable.
In the previous examples we did - "Derivatives by Limits" and "Derivatives by Rules", we dealt with explicit functions.
Here, we shall deal with implicit functions.
Say: $y=f(x)$
$y = x + 5 \rightarrow Explicit\:\: Function$
$y$ is expressed in terms of $x$

$y - x = 5 \rightarrow Implicit\:\: Function$
$y$ is not expressed in terms of $x$

Student: When we are given an implicit function, can we rearrange it to an explicit function like in the example you just gave, and find the derivative using the Power Rule?

Teacher: Yes, you can.
However, the example is a very simple example.
We shall do more challenging ones as we progress through the course.
It will be better and less time-consuming to differentiate implicitly rather than trying to convert it to an explicit function and differentiating explicitly.

Student: Do we have a whole new different set of rules for implicit differentiation?
It seems like it's a lot of rules.

Teacher: Good question. There are no new rules for implicit differentiation.
We shall learn a few more concepts/techniques and hence, have a broader understanding of the concept of derivatives. But, there are no new rules for this section.

### Notable Notes for Implicit Differentiation

(1.) Differentiate each term with respect to the independent variable.
In other words, differentiate each term $wrt\:\: x$

(2.) Differentiating $x\:\:wrt\:\:x = \dfrac{dx}{dx} = 1$

(3.) Differentiating $x^2\:\:wrt\:\:x = 2x\dfrac{dx}{dx} = 2x * 1 = 2x$

(4.) Differentiating $x^3\:\:wrt\:\:x = 3x^2\dfrac{dx}{dx} = 3x^2 * 1 = 3x^2$

(5.) Differentiating $y\:\:wrt\:\:x = \dfrac{dy}{dx}$

(6.) Differentiating $y^2\:\:wrt\:\:x = 2y\dfrac{dy}{dx}$

(7.) Differentiating $y^3\:\:wrt\:\:x = 3y^2\dfrac{dy}{dx}$

(8.) Differentiating $p\:\:wrt\:\:x = \dfrac{dp}{dx}$

(9.) Differentiating $p^2\:\:wrt\:\:x = 2p\dfrac{dp}{dx}$

(10.) Differentiating $p^3\:\:wrt\:\:x = 3p^2\dfrac{dp}{dx}$

(11.) Differentiating

$x^2y^3\:\:wrt\:\:x \\[3ex] = x^2 * 3y^2\dfrac{dy}{dx} + y^3 * 2x\dfrac{dx}{dx} ...Product\:\:Rule \\[5ex] = 3x^2y^2\dfrac{dy}{dx} + 2xy^3$

(12.) Use the Rules of Derivatives as applicable.

## Slope

Vocabulary Words
input, output, initial value, final value, change, increment, decrement, average rate of change, slope, secant line, intersects, passes, tangent line, touches, quotient, difference quotient, limit, derivative, point of tangency,

Recall:
Prior knowledge of PreAlgebra and Algebra
PreAlgebra: (Relations and Functions):
(1.) We defined slope as: ratio of the change in the output value of the function with respect to (wrt) a unit change in the input value of the function.
It is also known as the average rate of change

Considering a two-dimensional coordinate system where $y = f(x)$
$y$ = dependent variable
$x$ = independent variable

For a Linear Graph (Graph of a Linear Function) which is a straight line graph,

$Point\;1\;\;(x_1, y_1) \\[3ex] x_1 = initial\;\;value\;\;of\;\;x \\[3ex] y_1 = initial\;\;value\;\;of\;\;y \\[3ex] Point\;2\;\;(x_2, y_2) \\[3ex] x_2 = final\;\;value\;\;of\;\;x \\[3ex] y_2 = final\;\;value\;\;of\;\;y \\[3ex] Change = final\;\;value - initial\;\;value \\[3ex] Change\;\;in\;\;x = \Delta x = x_2 - x_1 \\[3ex] Change\;\;in\;\;y = \Delta y = y_2 - y_1 \\[3ex] Slope,\;\;m \\[3ex] = \dfrac{\Delta y}{\Delta x} \\[5ex] = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex]$ In this example, both changes are positive.
Hence, the slope is also positive.
As the value of $x$ increases, the value of $y$ also increases.
This implies that the rate of change of $y$ per unit change of $x$ is positive.

Algebra: (Difference Quotient):
(2.) We defined these terms:
(a.) The secant line to a curve is the line that intersects two points on the curve.

(b.) The tangent line to a curve is the line that touches only one point on the curve.

(Notice the terms: intersects, two points, touches, one point)

(3.) We reviewed the Difference Quotient of a function and noted these definitions.
We noted that:
(a.) The difference quotient of a function is the slope of the secant line.
(b.) The derivative of a function is the slope of the tangent line.

(4.) We also noted that for: the graph of all linear functions (straight-line graph): the slope of the line is the same at every point on the graph/line.
Depending on the graph, the slope can be positive, negative, zero, or undefined.

Calculus:
But, how do we find the slope of non-linear graphs (curves)?
For example, how do we find the slope of a: parabola? graph of a cubic function? graph of a quartic function? etc.

Well, here comes Calculus!
Because a non-linear graph is not a straight line, the slope is not the same at every point on the graph/curve.
The slope changes/varies (rises, falls, is constant) from one point to another on the graph depending on the intervals at which the graph is increasing, decreasing, or is constant.
In this case, the slope of the curve is not the same for every point on the curve.

Let us look at the case of a parabola (the graph of a quadratic function).

The slope of the three points on the curve are different because the graph is not a straight line. It is a curve.
The slope of point C on the graph is negative because the graph is decreasing at certain intervals in which point C is located (red color)
The slope of point B is zero because the graph is constant at the interval that has point B (yellow color)
The slope of point A is positive because the graps is increasing at some intervals that has point A (green color)
Notice that there are three tangent lines to that curve because each line touches the curve at only one point: Point A or Point B or Point C
The points (Points A, B, and C) are known as points of tangency
The point of tangency is the point of intersection of the tangent line and the curve.
So, the question is: how can we find the slope of any point say point A on a curve?

Finding the Slope of a Curve at a Point of Tangency
(1.) First Approach: Visual Approximation: One way to do this is to approximate it.
This approach involves taking some measurements of the input ($x$) and the output ($y$) value that contains the point of tangency.
With this approach, it is important to note the scale of the graph on both axis: the $y-axis$ (horizontal axis) and the $x-axis$ (vertical axis)

$Slope\;\;of\;\;the\;\;Curve\;\;at\;\;Point\;A \\[3ex] m = \dfrac{\Delta y}{\Delta x} \\[5ex]$ (2.) Second Approach: Difference Quotient
This approach uses a secant line.
It is a more precise approach of approximating the slope of a curve at the point of tangency.
For a secant line, two points are needed.
The first point is the point of tangency.
The second point is another point on the curve.
Remember that we want to find the slope of the curve at Point A
Point A becomes out first point: Point 1
We find another point on the curve and call it Point 2

$Slope\;\;of\;\;the\;\;Curve\;\;at\;\;Point\;A \\[3ex] Point\;A = Point\;1 = (x_1, y_1) \\[3ex] Point\;2 = (x_2, y_2) \\[3ex] y = f(x) \\[3ex] \Delta x = x_2 - x_1 \\[3ex] x_2 - x_1 = \Delta x \\[3ex] x_2 = \Delta x + x_1 \\[3ex] x_2 = x_1 + \Delta x \\[5ex] Let:\;\; x_1 = x \\[3ex] \implies \\[3ex] y_1 = f(x) \\[3ex] x_2 = x + \Delta x \\[3ex] y_2 = f(x_2) \\[3ex] y_2 = f(x + \Delta x) \\[3ex] Slope,\;\;m \\[3ex] = \dfrac{\Delta y}{\Delta x} \\[5ex] = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{f(x + \Delta x) - f(x)}{(x + \Delta x) - x} \\[5ex] = \dfrac{f(x + \Delta x) - f(x)}{x + \Delta x - x} \\[5ex] = \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \\[5ex] Difference\;\;Quotient = DQ \\[3ex] DQ = \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \\[5ex]$ This is the Slope of the secant line
It is the average rate of the function at point A (point of tangency)
It is also known as the Difference Quotient because it is a quotient of two differences.
The numerator is a difference.
The simplified denominator was a difference.

Here is the interesting thing:
The second point we chose is a bit distant.
We can choose a point very close to the first point (the point of tangency)
By choosing points closer and closer to the point of tangency (the first point), more accurate approximations of the slope of the tangent line is obtained.
This implies that as the secant line approaches the tangent line, more accurate approximations of the slope is obtained.
This implies that as the change in $x$: $\Delta x$ approaches zero, more accurate approximations of the slope is obtained.
Hence, the definition of limit is introduced.

(3.) Third Approach: Derivative
This approach uses the limit of the difference quotient as the change in the independent variable approaches zero.
It gives the exact value of the slope of the curve at the point of tangency.

$\displaystyle{\lim_{\Delta x \to 0}} DQ \\[5ex] = \displaystyle{\lim_{\Delta x \to 0}} \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \\[5ex] = \dfrac{dy}{dx} \\[5ex] \therefore \dfrac{dy}{dx} = \displaystyle{\lim_{\Delta x \to 0}} \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \\[5ex] \dfrac{dy}{dx} = f'(x) \\[3ex] So: \\[3ex] f'(x) = \displaystyle{\lim_{\Delta x \to 0}} \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \\[5ex]$ This is the Slope of the tangent line
It is the instantaneous rate of the function at Point A (the point of tangency)
It is also known as the Derivative
This gives us the exact value of the slope of the tangent line to the curve at Point A (the point of tangency)
It is the slope of the graph at $(x, f(x))$

Everything must not be $x$
We can use any variable as the independent variable.
Assume we decide to use the variable: $a$ rather than $x$, we have:

$DQ = \dfrac{f(a + \Delta a) - f(a)}{\Delta a} \\[5ex] f'(a) = \displaystyle{\lim_{\Delta a \to 0}} \dfrac{f(a + \Delta a) - f(a)}{\Delta a} \\[5ex]$ The difference quotient is the approximate slope of the graph at $(a, f(a))$
The derivative is the exact slope of the graph at $(a, f(a))$

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