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

Prerequisite 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

Prerequisite Knowledge (from the Prerequisite 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.

### References

Chukwuemeka, S.D (2016, April 30). Samuel Chukwuemeka Tutorials - Math, Science, and Technology. Retrieved from https://www.samuelchukwuemeka.com

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Stroud, K. A., & Booth, D. J. (2001). Engineering Mathematics ($5th$ ed.). Basingstoke: Palgrave.

Tan, S. T. (2004). Applied Calculus for the Managerial, Life, and Social Sciences ($5th$ ed.). Pacific Grove, CA: Brooks/Cole Publishing Company.

Waner, S., & Costenoble, S. (2013). Applied Calculus ($6th$ ed.). Boston, MA: Cengage Learning.

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School Curriculum and Standards Authority (SCSA): K-12. Past ATAR Course Examinations. Retrieved December 10, 2020, from https://senior-secondary.scsa.wa.edu.au/further-resources/past-atar-course-exams

West African Examinations Council (WAEC). Retrieved May 30, 2020, from https://waeconline.org.ng/e-learning/Mathematics/mathsmain.html