Implicit Differentiation Explained - Product Rule, Quotient & Chain Rule - Calculus
Summary
TLDRThis 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
- ๐ Implicit differentiation is used when you can't easily solve for y in terms of x.
- ๐ข Differentiate both sides of an equation with respect to x to find dy/dx.
- โ๏ธ When differentiating y terms, multiply by dy/dx.
- ๐ The derivative of a constant is zero.
- ๐ Isolate dy/dx by moving terms without dy/dx to the other side of the equation.
- โ To solve for dy/dx, divide both sides of the equation by the coefficient of dy/dx.
- ๐ Use the product rule when differentiating terms involving both x and y.
- ๐ For the second derivative, use the quotient rule if dealing with a fraction.
- ๐ When differentiating trigonometric functions, use their respective derivative rules.
- ๐ 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
๐ 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.
๐ 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.
๐งฎ 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.
Mindmap
Keywords
๐กImplicit Differentiation
๐กDerivative
๐กProduct Rule
๐กChain Rule
๐กQuotient Rule
๐กSecant Square
๐กIsolating dy/dx
๐กDifferentiation of Constants
๐กSimplifying Equations
๐กSecond Derivative
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
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