This calculus video tutorial explains the concept of implicit differentiation and how to use it to differentiate trig functions using the product rule, quoti. 6.3 Explicit Vs Implicit Differentiation Notes No tes Key. Homework Hw Key. Powered by Create your own unique website with customizable templates. Through the implicit function theorem, we are guaranteed that y can be represented as a single variable function of x, this helps a lot with finding that limit.
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In many examples, especially the ones derived from differentialequations, the variables involved are not linked to each other inan explicit way. Most of the time, they are linked through animplicit formula, like F(x,y) =0. Once x is fixed, we mayfind y through numerical computations. (By some fancy theorems,we may formally show that y may indeed be seen as a function ofx over a certain interval). The question becomes what is thederivative ,at least at a certain apoint? The method of implicit differentiation answersthis concern. Let us illustrate this through the followingexample.
Example. Find the equation of the tangent line to theellipse
at the point (2,3). One way is to find y as a function ofx from the above equation, then differentiate to find the slopeof the tangent line. We will leave it to the reader to do thedetails of the calculations. Here, we will use a differentmethod. In the above equation, consider y as afunction of x:
and use the techniques of differentiation, to get
From this, we obtain
which implies that at the point (2,3). So the equation of the tangent line isgiven by
You may wonder why bother if this is just a different way offinding the derivative? Consider the following example! It can bevery hard or in fact impossible to solve explicitly for y as afunction of x.
Example. Find y' if
This is a wonderful example of an implicit relation between xand y. So how do we find y'? Let us differentiate the aboveequation with respect to x where y is considered to be afunction of x. We get
Easy algebraic manipulations give
We can also find higher derivatives of y such as y' in thismanner. We only have to differentiate the above result. Ofcourse the calculations get little more messy.
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Exercise 1. Find y' if xy3 + x2y2 + 3x2 - 6 = 1.
Exercise 2. Prove that an equation of the tangent line tothe graph of the hyperbola
at the point P(x0,y0) is
Exercise 3. Show that if a normal line to each point on anellipse passes through the center of an ellipse, then the ellipseis a circle.
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Integration Mini Video Lecture
This video explains the differences and similarities between indefinitie and definite integration.
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This mini video lecture explains the processes in these two pages:
... and ...
What's the difference between indefinite and definite integrals?
With an indefinite integral there are no upper and lower limits on the integral here, and what we'll get is an answer that still has x's in it and will also have a K, plus K, in it.
A definite integral has upper and lower limits on the integrals, and it's called definite because, at the end of the problem, we have a number - it is a definite answer.
OK. Let's do both of them and see the difference.
The integral x to the five dx is equal to ... `(int x^5 dx =?)`
... Now, for integration, I have to add one to the index.
So, it's going to be x to the 6
...and then I divide by the new number ... x to the sixth over 6 `(int x^5 dx = x^6/6...)`
...and then I must remember the plus K. `(int x^5 dx = x^6/6+K)`
K is just some constant. We might have some information elsewhere in the problem that will help us to find the constant. In this case we don't have any extra information, so it's just plus K.
Now, let's see what it looks like as a definite integral, this time with upper and lower limits, and we'll see what happens. `(int_1^2 x^5 dx = ?)`
Step 1 is to do what we just did. We integrate, and I'm going to have once again x to the six over 6, but this time I do not have plus K - I don't need it, so I don't have it.
And I write it like this,
... and then I have a lower limit of 1
... and an upper limit of 2. `(int_1^2 x^5 dx = [x^6/6]_1^2)`
These are the numbers that come from here in the question - 1 comes from here, 2 comes from here.
So, what I've done is I've integrated, and ... written in the upper and lower limits correctly.
The next step is to actually substitute these upper and lower limits.
So, what I can do is like this ... and like this ... and I'm going to substitute 2 into this position, and 1 into this position.
So, what have I done? `(int_1^2 x^5 dx = [x^6/6]_1^2=2^6/6-1^6/6)`
I've found the integral, and I've substituted 2 into this position ... and 1 into this position ... and I'm going to subtract the difference ... and I'm just going to use Scientific Notebook now to give me that, as a decimal.
6.1 Implicit Vs Explicitap Calculus Algebra
And it turns out to be 10.5.
`int_1^2 x^5 dx = [x^6/6]_1^2` `=2^6/6-1^6/6` `=10.5`
And this is a definite answer.
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