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Open original 5Give the equation V2(I of the tangent line 3 urGsin 11A at the point 1...

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Open original 5Give the equation V2(I of the tangent line 3 urGsin 11A at the point 1

Open original 5 Give the equation V2(I of the tangent line 3 urGsin 11A at the point 1



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Find the equation of the tangent line at the indicated point (see Example 3 ). $$x^{3} y+y^{3} x=30 ;(1,3)$$

Okay, so this question wants us to great the equation of the tangent line to a curve, given these two pieces of information. So we know a point on the graph and we know the tangent line slope at that point. So remember, we can write a line in the form. Why minus why not equals m times X minus x Not where that's not. Why not is a point on the grass. So we know the tangent line hits the curve at the point of tendency, which is our first order pair here. 35 And we also know the slope at that point, which is to given by the derivative. So now we can plug in why minus five equals two times X minus three. Now, if you're on the AP exam, you can just leave this as is, and you'll get full credit. But we usually like lines in our y equals MX plus B form. So why minus five equals two X minus six? And then we had five to both sides to get why equals two X minus one, and that's our equation of the tangent line.

So whenever we want to find the equation of a tangent line, it's important to think about how do you write the equation of a line. What's most helpful is the point slope form of a line, which is why minus Y. One equals M X minus X. One. And we already know a given why value is 1/5. And for that why we know X is one. So really all we need to do is find the slope and that instantaneous slope is called a derivative. So how are we going to find the derivative of our function? Well, since it's a quotient, something divided by something else, let's use the quotient rule. So the derivative which I'll denote as Y prime is the derivative of the numerator times the denominator unchanged and minus the derivative. The denominator with the numerator unchanged. And then all divided by that denominator but squared, Well the derivative of one is 0. So that eradicates that first term and the derivative of X squared plus four would be two X plus zero. So we get negative two x times one over X squared plus four. But squared When we plug in our known x value of one We get negative two times 1 In one squared plus four is 5 but squared is 25 and that's our slip. So going back to our point slope form, we know the equation of our line are tangent line is why -1 5th M Now is negative 2 25th An X -1. So with the quotient rule in the point slope form of the line were able to find the equation of the tangible. Um

In the problem we have three X Squire. Why plus X. Q. White ass plus Y. Que plus X. into three. Why Squire? Why does that equal zero now XQ plus three X. Or the Squire? Where does become -3? Expo square wine minus? Like you r minus equal to minus three x. esquire y less. Like you bond X cubed plus three X. Y. Square. So we have, why does add one Common 3 equal to -9 Upon 7? Hence the equation. Uh huh. Detainment line become why am minus? Why not? So why not? is three That equals m. Which is -9.7 X -X Not is one. So this is the occasion of dependent line, hence it is dancer.

Hello. So here we have the function G f x is equal to two X minus one over x minus four. And I'm gonna find the derivative G. Prime of X. Using the definition. So the definition here is going to be to the limit as H goes to zero of well G. Of X plus H minus G F X all divided by H. So that's going to be the limit as H go to zero of two times X plus H um minus one over um X plus h minus four. Just putting in an X plus H wherever we see an X. And our function. And then we're subtracting off G. F. X. That subtracting off two x minus one over x minus four. And then we are all dividing this whole thing by H. So we just get a common denominator now to combine the numerator here. Um So that's going to be a X plus H minus four times X minus four. Um So what we end up with here is, let's see, we have still the limit as H coasters. They're all of, we then have a two X plus two H minus one times x minus four minus two X minus one times X plus h minus four. Uh And then that is going to be divided by the common denominator here of X plus H minus four times X minus four. And then we see that then is all divided by H. Okay, so well we just go ahead, we distribute here and then we have a minus sign to distribute and then what we end up with, I'm gonna write it down, I have enough room here but we have to limit as H ghost zero of we have a two X squared plus two X h minus x minus eight X minus eight, H plus four minus two, X squared minus two X h plus eight X plus X plus h minus four. So that whole thing is the numerator and that's over the same common denominator of the X plus H minus four times x minus four. And again that's all divided by H. But then there's a bunch of cancellations and the only thing that we're left with on top is going to be um a negative eight H plus H. That's going to be a negative seven H. So then we have the limit as a church goes to zero of a negative seven H. Um divided by well X plus H minus four times x minus four over age which is times one over age. So therefore we have native +78 times each times X plus H minus four times x minus four. Now the H over H here is going to cancel out. And then we actually we can actually take the limit let h go to zero. So we have X plus zero minus four. So that is going to go away and we're just left with well negative seven over x minus four times x minus four, so x minus four times x minus four, but that's just x minus four squared. So before we see that the derivative here um is going to be equal to negative seven over the quantity x minus four squared, so, mhm.


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