Implicit Differentiation Explained - Product Rule, Quotient & Chain Rule - Calculus

Summary

This video explains the process of implicit differentiation through various examples. Starting with basic equations like x^3 + y^3 = 8, the instructor shows how to differentiate both sides with respect to x, while applying the chain rule when dealing with y. The video also covers how to isolate dy/dx, use product and quotient rules, and simplifies expressions to make the process clearer. With detailed walkthroughs, the instructor guides viewers in solving more complex problems, enhancing their understanding of implicit differentiation techniques.

Takeaways

• 📘 Implicit differentiation is used when you can't easily solve for y in terms of x.
• 🔢 Differentiate both sides of an equation with respect to x to find dy/dx.
• ✏️ When differentiating y terms, multiply by dy/dx.
• 🔄 The derivative of a constant is zero.
• 🔄 Isolate dy/dx by moving terms without dy/dx to the other side of the equation.
• ➗ To solve for dy/dx, divide both sides of the equation by the coefficient of dy/dx.
• 🔄 Use the product rule when differentiating terms involving both x and y.
• 🔄 For the second derivative, use the quotient rule if dealing with a fraction.
• 📐 When differentiating trigonometric functions, use their respective derivative rules.
• 🔄 Always check if you can simplify the equation before differentiating to make the process easier.

Q & A

What is the process of implicit differentiation?

Implicit differentiation is a technique used to find the derivative of a function that is not expressed explicitly in terms of y. It involves differentiating both sides of an equation with respect to x, treating y as an implicit function of x, and then solving for dy/dx.

How do you differentiate the equation x^3 + y^3 = 8 with respect to x?

To differentiate x^3 + y^3 = 8 with respect to x, you differentiate each term: the derivative of x^3 is 3x^2, and for y^3, you use the chain rule, resulting in 3y^2 * dy/dx. After differentiating, you get 3x^2 + 3y^2 * dy/dx = 0. Then, isolate dy/dx to find that dy/dx = -x^2/y^2.

What is the product rule in the context of implicit differentiation?

The product rule in implicit differentiation is used when differentiating a product of two functions, such as x*y. It states that the derivative of the product is the derivative of the first function times the second function plus the first function times the derivative of the second function. In implicit differentiation, this often involves adding dy/dx to one of the terms.

How do you find dy/dx for the equation x^2 + 2xy + y^2 = 5?

Differentiating x^2 + 2xy + y^2 = 5 with respect to x, you get 2x + 2y(dy/dx) + 2y(dy/dx) = 0. Isolating dy/dx, you move terms without dy/dx to the other side and factor out dy/dx, resulting in dy/dx = -2x / (2y + 2x), which simplifies to dy/dx = -x / (y + x).

What is the derivative of a constant in implicit differentiation?

The derivative of a constant in implicit differentiation is always zero, as constants do not change with respect to the variable being differentiated.

Can you provide an example of how to solve for dy/dx when the equation involves a trigonometric function?

For the equation tan(xy) = 7, you differentiate using the chain rule, resulting in sec^2(xy) * (x + y * dy/dx) = 0. Isolating dy/dx, you distribute sec^2(xy) to x and y * dy/dx, and then solve to find dy/dx = -y/x.

What is the quotient rule in calculus, and how is it used to find the second derivative?

The quotient rule is used to find the derivative of a fraction, where the numerator and denominator are both functions of x. It states that the derivative of u/v is (v*u' - u*v') / v^2. To find the second derivative, you apply the quotient rule to the expression for dy/dx, treating it as a fraction where u and v are the numerator and denominator, respectively.

How do you differentiate the equation 5xy - y^3 = 8 with respect to x?

Differentiating 5xy - y^3 = 8 with respect to x, you get 5 + 5y(dy/dx) - 3y^2(dy/dx) = 0. Isolating dy/dx, you factor out dy/dx and solve to find dy/dx = -5y / (5x - 3y^2).

What is the significance of simplifying an equation before differentiating it?

Simplifying an equation before differentiating it can make the process easier by reducing the complexity of the terms involved. For example, squaring both sides of an equation can eliminate square roots, making differentiation simpler and the resulting expressions easier to manage.

How do you find the second derivative of dy/dx = x/y using the quotient rule?

To find the second derivative of dy/dx = x/y, you apply the quotient rule to the expression for dy/dx. Let u = x and v = y, then u' = 1 and v' = dy/dx. Plugging into the quotient rule formula, you get d^2y/dx^2 = (y * 1 - x * dy/dx) / y^2, and substituting dy/dx = x/y, you simplify to find the second derivative.

Outlines

00:00

📚 Introduction to Implicit Differentiation

This paragraph introduces the concept of implicit differentiation, a method used to find the derivative of a function that is not explicitly solved for y. The video explains how to differentiate both sides of an equation with respect to x, handling terms involving y by adding dy/dx to them. An example is given where the function x^3 + y^3 = 8 is differentiated to find dy/dx, resulting in dy/dx = -x^2/y^2 after isolating dy/dx. The process involves moving terms without dy/dx to the other side of the equation and simplifying.

05:03

🔍 More Examples of Implicit Differentiation

The paragraph presents additional examples to further illustrate implicit differentiation. The first example involves the equation x^2 + 2xy + y^2 = 5, where the product rule is applied to differentiate x and y. After differentiating and rearranging terms, dy/dx is found to be 1. The second example uses the equation 5xy - y^3 = 8, applying the product rule and isolating dy/dx to get dy/dx = -5y/(5x - 3y^2). The video emphasizes the importance of moving terms without dy/dx to one side and factoring out dy/dx to solve for it.

10:07

🧮 Advanced Implicit Differentiation Scenarios

This section tackles more complex scenarios in implicit differentiation, including the use of trigonometric functions and the chain rule. An example with the equation tan(xy) = 7 is differentiated to find dy/dx, leading to two possible solutions: dy/dx = -y/x or dy/dx = -y/x, depending on the approach taken. The video also covers a scenario with the equation 36 = x^2 + y^2, where squaring both sides simplifies the differentiation process, resulting in dy/dx = x/y. The paragraph concludes with a brief introduction to finding the second derivative using the quotient rule.

Keywords

💡Implicit Differentiation

Implicit differentiation is a technique used in calculus to find the derivative of a function that is not expressed explicitly in terms of y. It's a method where the derivative \( \frac{dy}{dx} \) is found by differentiating both sides of an equation with respect to \( x \), treating \( y \) as an implicit function of \( x \). In the video, this concept is central to solving problems where \( y \) is not isolated, such as in the equation \( x^3 + y^3 = 8 \). The process involves differentiating both sides of the equation and then isolating \( \frac{dy}{dx} \).

💡Derivative

A derivative in calculus represents the rate at which a function changes with respect to its variable. It is the slope of the tangent line to the graph of the function at a given point. The video script uses derivatives to find the rate of change of \( y \) with respect to \( x \) in various equations, such as \( x^2 + 2xy + y^2 = 5 \), showcasing how to apply differentiation rules to both sides of an equation.

💡Product Rule

The product rule is a differentiation rule used when differentiating a product of two functions. It states that the derivative of a product is the derivative of the first function times the second function plus the first function times the derivative of the second function. In the video, the product rule is applied when differentiating terms involving both \( x \) and \( y \), such as in the equation \( 5xy - y^3 = 8 \), to find \( \frac{dy}{dx} \).

💡Chain Rule

The chain rule is a fundamental calculus rule used to differentiate composite functions, which are functions composed of other functions. It states that the derivative of the composite function is the derivative of the outer function times the derivative of the inner function. In the video, the chain rule is mentioned in the context of differentiating \( \tan(xy) \), where the derivative of \( \tan \) is \( \sec^2 \), and the inner function \( xy \) is differentiated accordingly.

💡Quotient Rule

The quotient rule is a calculus rule used to differentiate a quotient of two functions. It states that the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. The video script refers to the quotient rule when finding the second derivative of \( \frac{x}{y} \) from the equation \( x^2 + y^2 = 36 \), illustrating how to handle fractions when taking derivatives.

💡Secant Square

Secant squared, denoted as \( \sec^2 \), is a trigonometric function that appears in the derivative of the tangent function. It is equal to \( 1 + \tan^2 \). In the video, secant squared is used in the differentiation of \( \tan(xy) \), where the derivative of \( \tan \) is \( \sec^2 \), and the script explains how to handle the differentiation of the inner function \( xy \).

💡Isolating dy/dx

Isolating \( \frac{dy}{dx} \) is a process in differentiation where the goal is to express the derivative \( \frac{dy}{dx} \) by itself on one side of the equation. This is crucial in implicit differentiation to solve for the rate at which \( y \) changes with respect to \( x \). The video provides examples of how to rearrange terms and simplify equations to isolate \( \frac{dy}{dx} \), such as in the equation \( x^3 + y^3 = 8 \).

💡Differentiation of Constants

In calculus, the derivative of a constant is always zero. This is because constants do not change with respect to the variable. The video script illustrates this concept when differentiating equations involving constants, such as in \( x^3 + y^3 = 8 \), where the derivative of the constant 8 is zero, simplifying the differentiation process.

💡Simplifying Equations

Simplifying equations is a common technique used in mathematics to make complex expressions more manageable. In the context of the video, simplifying equations is shown as a strategy to make differentiation easier, especially before applying differentiation rules. An example from the script is squaring both sides of \( x^2 + y^2 = 36 \) to eliminate the square root and simplify the differentiation process.

💡Second Derivative

The second derivative is the derivative of the first derivative of a function. It measures the rate of change of the first derivative and can provide information about the concavity of a function's graph. In the video, the second derivative is calculated for the function \( \frac{x}{y} \) derived from \( x^2 + y^2 = 36 \), using the quotient rule to find \( \frac{d^2y}{dx^2} \).

Highlights

Introduction to implicit differentiation

Differentiating both sides of the equation with respect to x

Derivative of x^3 is 3x^2 and y^3 is 3y^2 * dy/dx

Differentiating a constant results in zero

Isolating dy/dx by moving terms to the other side

Solving for dy/dx by dividing both sides by 3y^2

Differentiating x^2 + 2xy + y^2 = 5 using the product rule

Separating the function into two parts for product rule application

Derivative of y^2 is 2y * dy/dx

Isolating dy/dx by moving terms without dy/dx to the other side

Simplifying the equation by dividing everything by 2

Factoring out dy/dx to solve for it

Solving for dy/dx in the equation 5xy - y^3 = 8

Using the product rule for differentiating 5x * y

Isolating dy/dx by moving terms without dy/dx to the other side

Solving for dy/dx by dividing both sides by 5x - 3y^2

Differentiating tangent(xy) = 7 using secant squared

Applying the chain rule to differentiate the inside of tangent

Isolating dy/dx by distributing secant squared to y and x * dy/dx

Simplifying to find dy/dx as negative y over x

Alternative method for solving the tangent(xy) = 7 problem

Squaring both sides to simplify the derivative of x^2 + y^2 = 36

Differentiating both sides with respect to x and simplifying

Finding the second derivative using the quotient rule

Applying the quotient rule to find d^2y/dx^2

Final solution for the second derivative

Transcripts

in this video we're going to talk about

how to do implicit differentiation so

let's say if you have this function X

Cub

plus let's say y Cub is equal to 8 and

you want to find uh

dydx at let's say you want to just find

dydx so what we need to do is

differentiate both sides with respect to

X so that's D over DX the derivative of

x 3r is

3x^2 and the derivative of y 3 is 3 y^2

* dydx for these kinds of problems

anytime you differentiate a y value add

dydx to it the derivative of a constant

like a is zero so now we got to put dydx

we got to get it by itself so if we move

the 3x s to the other side here's what

we now

have all we got to do now is divide both

sides by 3

y^2 and so in this particular case dydx

is therefore equal to the 3es cancel

it's- x^2 over

y^2 so that's how you can do it but now

let's try um another

example so let's say

if we have this function x^2 + 2x y +

y^2 = 5 and we want to find

dydx so let's go ahead and jump right

into it let's differentiate x^2 to

derivative of X2 is 2x now for this part

right here we have X and Y combined and

whenever you see that you need to use

the product rule so let's separate this

function into two parts 2X and time

y so to use the product rule the gist of

it is like this you differentiate one

part and then keep the other part the

same so we're going to differentiate the

2x part which is going to become two

we're going to keep y the same plus we

need to keep the first part the same now

and then differentiate the second part

the derivative of y is one and anytime

you differentiate a y variable in

relation to implicit differentiation add

dydx to

it so here we have y^ 2 next and the

derivative of y^2 is 2 y *

dydx and the derivative of any constant

is always

zero so now our goal right now is to

isolate dydx so any variable that or any

term that doesn't have a dydx we're

going to move it to the right side so

what we now have on the left is 2x Dy

DX plus 2 Y dydx is equal to -2X - 2 y

there's a lot of twos here

you know what let's divide everything by

two so this will disappear that and that

all of them will

disappear our next step we're going to

factor out

dydx so we're left with X + Y is equal

tox minus y which we're going to factor

out negative 1 and that's x +

y so now we're going to divide both

sides by x +

y so notice that the X plus y cancel so

for this particular problem dydx is just

equal

to1 let's try another problem for the

sake of

practice so let's say if you have

5xy minus y 3r is equal to 8 now feel

free to pause the video give this

problem a shot and find

dydx so let's use the product rule here

so the first part it's going to be 5x

the second part is y the derivative of

5x is simply five and then times the

second part

Y and we're going to now we're going to

keep the first part the same 5x and then

we're going to differentiate Y which is

1 * Dy

DX the derivative of y 3 is 3 y^2

dydx and the derivative of a constant is

zero

so any term that doesn't have a dydx

let's move it to the other side so

that's the 5y we're going to move it

over

here so now we have 5x

dydx - 3 y^2 dydx is equal to - 5 y

whenever you move a term from one side

to another it changes from positive to

negative or vice versa our next step is

to factor out the GCF well not really

the GCF but just dydx we need to isolate

it so we have dydx

time 5x - 3 y^

2 = -5

y so our last step is to get dydx by

itself by dividing both sides by 5x - 3

y^2 and so now we have our answer dydx

is equal to

-5 y over 5x - 3

y^2 so that's it for that

problem but there's more to talk

about so let's say if you have this

problem let's

say

tangent

XY is equal to

7 go ahead and find uh dydx for this

problem

the derivative of tangent is secant

squar and we got to keep the inside the

same the angle for tangent was XY so

therefore the angle for secant squ has

to be XY but now we got to differentiate

the inside part according to the rules

of the chain rule so let's use the

product

rule the derivative of x is one keep the

second part the same the derivative of y

Y is 1 * dydx but we got to keep the

first part the same which is the

X and the derivative of a constant is

zero now for this problem to get dydx by

itself the first thing we need to do is

distribute this term secant squ to Y and

to X

dydx so secant squ * Y is simply y

secant s

XY and for the next term this time see s

is just X

dydx time see 2

XY so now what we're going to do is um

we're going to take this term which

doesn't have a dydx and we're going to

move it to this

side so X dydx secant

squ XY is equal to y^ 2 XY

you know I realized this probably was an

easy way of doing this which I'm going

to show you after this part let's divide

both sides by X secant

squ notice that the secant squar is

cancel and in the end dydx is basically

negative y overx so but let me show you

the other way we can have done this

so we had see 2

XY time I believe it

was y + x dydx is equal to

Zer what we could have done at this

point is just at this right here divide

both sides by secant

Square so these cancel and you get y + x

dydx is equal to0 0 divided by anything

is zero and yeah it's going to be much

easier solving it like this so now let's

move y to the other

side and then let's divide both sides by

X so therefore

dydx is equal to-

y/x so there's more than one way of

solving a

problem let's try another one let's say

if 36 is equal to theun of x^2 +

y^2 how would you find the answer for

this

one now you can go ahead and take the

derivative at this point but personally

I believe it's much easier if you

simplify it or adjust it let's Square

both

sides we really don't need to find out

what 36 squ is we're just going to write

it as 36 squ because when we take the

derivative it's still going to be a

constant and it's going to become zero

however this part is very useful when

you square a square root the radical

disappears and it makes it a lot easier

to find the derivative so now let's

differentiate both sides with respect to

X the derivative of 36 SAR which we

already said was a constant is zero the

derivative of x^2 is 2X and for y^2 it's

2 y * dy/ DX so let's move the 2 X to

this side it's going to become

-2X and then let's divide both sides by

2

y so the twos cancel so therefore dy

over DX is thus equal

tox over

Y at least for this particular

example and let's say though we have to

find the second

derivative so we have this equation dydx

X is equal

tox y whenever you want to find a second

derivative and since we have a fraction

we need to use the quotient

rule which is v u Prime minus UV Prime

over V

^2 U is basically the top part which

ISX U Prime actually let me change the

color so U

ISX U Prime is the derivative of

Negative X which

is1 uh V is the bottom part which is y

and V Prime is the derivative of y which

is 1 *

dydx so let's use the formula so D ^2 y

over dx^ 2 the second derivative is

equal to V which is

y time U Prime which is

-1 minus

U which

ISX time V

Prime which is just dy over

DX okay so and

v^2 is simply

y^2 notice that we have a dydx here what

we need to do is take this term and plug

it in for dydx so we have- y + x * X

over y time y^2 now what we're going to

do is multiply top and bottom by y just

to get rid of this uh mini

fraction so- y * y see if I can fit that

here y^

2 and the Y's cancel here so we get x^2

over whatever you do to the top you have

to do to the bottom so you got to

multiply top and bottom by y so over y

Cub that's the second derivative for

this function that's how you could find

it so that's it for this video um

hopefully this gave you more or better

Insight on how to find uh the derivative

us an implicit differentiation