8. (3 points) Find the equation of the tangent line to the graph of y In(x2 + e) atx = 0.0-00D0it(3 points) Solve for...

Find the slope of the tangent line to the graph at the given point.
Folium of Descartes:
$x^{3}+y^{3}-6 x y=0$
Point: $\left(\frac{4}{3}, \frac{8}{3}\right)$

So how are we going to get started with this problem? We need a derivative to make sense of the slope of the tangent line. So let's calculate the derivative of why with respect to X. That would be each of the 3X. But times 3 3 due to the chain rule, the derivative of the power itself. So let's imagine that the point on the grass, is it some X. Value A. And the output would then be E. to the three a. So now we know the slope at that point would be well we just plug in A. It would be three E. 23 A. So now we can construct the tangent line equation at any point. We'll use the point slope form. So that's why minus the known why value equals the slope X minus the known X. Value. But we know that we want 00 to be on the line. So everywhere. Um Y and X. Are we could substitute zero. Can I get something like this? We see a negative sign can cancel. We see the eat of the three a. Since it's not zero could cancel and that gets us the A. Is a third. So uh what we can then do is plug that back into our point. And the point of tangent see that we need is one third and then E. To the one. Yeah. So that's the point of tangent C. On that graph where its tangent line will pass through the origin.

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.

Were given a tangent line problem. And whenever you have a tangent line problem, just make sure that you find the point and they give you the ordered pair of the point. Which is generous because you don't need, you only need the X coordinate to be able to solve the problem. Um They go ahead and figure out the why coordinate for you. And then also uh we need to find the slope which is the derivative at that ordered pair. So in this problem we only need to know X equals zero as I mentioned before. So they give to us this uh composition function he to the two x plus one. And that quantity is cubed. And notice I called it the composition function because you have an inner function with an outer function. So when you go to find this derivative you start with the derivative of the outside Which would be three something squared. And what you do is you leave the inner function alone. But then you have to multiply by the derivative of the inside of, switch back to red, which is actually another chain rule. But we've been practicing it in this problem where the derivative of each of the two exes itself times the derivative of that experiment which should be too. Um and then the drift of a. one is zero. So we don't have to write that down. But we're ready to solve this because we can plug zero in for all of these exes. Um And just so we're clear either the zero power is equal to one, so we have one plus one is two And E to the zero power is another one times that too. So we have two squared is four times three is 12 times two is 24. And uh you could actually just do point slope form. I'm a big fan of that for your tangent line. Um, the slope and an X minus the X coordinate plus the Y coordinate in case you forgot zero and eight or the X and Y. Corn. So this is a good answer. However, you might have a teacher look at you and say, why did you write minus zero? Because x minus zero is just X. Um, so we're left was just 24 X plus eight, uh, insult ownership form. So we're good.

So we're gonna be finding all the points for this function F. Of X, whose slope is perpendicular to this function five Y minus three X. Is equal to eight. So let's go ahead and get this function into slope intercept form to do that. We're gonna want to plus three X over. So we have five. Y. Is equal to three X plus eight, divide both sides by five And we get y equal to 3/5 times x plus eight divided by five. And so the slope here is going to be equal to this 3/5. And so we want to slope that's perpendicular to that. So that would be the negative reciprocal or negative 5/3. And so what we want to do now is we want to find um f prime of X. Which we can find just using the power rule. So for this first one X cubed and would be equal to three. So the derivative equal to three X squared. And then here for negative three X. And is equal to one. So we're gonna have minus three. And so we just want to set this equal to negative 5/3 and then solve for X. So we have negative five thirds is equal to three times X squared minus three plus three over. So three is nine thirds, so nine thirds plus ar minus five thirds would be four thirds. So we have 4/3 is equal to three X squared, divide both sides by three. We get four. Nineths is equal to x squared square root both sides. And then we get the square to four which is to divided by the square root of nine which is three. So X. Here is going to be equal to two thirds. And now we just want to plug in two thirds into our original equation to find the point at which are um function has a tangent line perpendicular to um Why is equal to 3/5 X plus 8/5. So We're looking at f of 2/3 is equal to um X. Cube. So 2/3 Cubed and then -3 times X. So two thirds cubed is two cubed, which is eight Divided by three Cubes, which is 27 And then-3 times 2/3 would be -2 and two is going to be equal to 54 27. So this is eight, 27th minus 54 27th. Just going to be equal to negative 46 27th. And so the point that is perpendicular which has a tangent line that is perpendicular to the line that we're looking at is two thirds comma negative 46 27th.