![]() ![]() And let's see, I could subtract oneĮighth from both sides. And so let's see, I have one eighth plus two times the second derivative is equal to one fourth. Plus two times the second derivative of y with respect to x is equal to one fourth. One fourth is six fourths, minus one is going to be, so six fourths minus fourįourths is two fourths, so this is going to be one half times one fourth, times one fourth, that's this over here, plus the second derivative of y with respect to x, and here I have one minus negative one, so it's one plus one, so times two. So if six times one times one fourth is going to be one point five, or six, well, six times one times And now, we can solve for the second can solve for the second derivative. So this is one fourth, this is one fourth, and this is one fourth. Yep, at the point negative one comma one. Are there any other xs here? And what's dy dx? Well, dy dx, dy dx is going to be equal to one over three times one, which is three, minus negative one. So that's going to be equal to one and this is going to be equal to one. So, we know that, what we wanna figure out, our second derivative, when y is equal to one. Second derivative of y with respect to x, but what could be even better than that is if I substituteĮverything else with numbers because then it's just going to be a nice, easy numerical equation to solve. ![]() That's going to be equal to, on the right-hand side, derivative of y with respect to x is just dy dx. Once again, I'll say it a third time, derivative of this times this plus the derivative of this times that. Well, that's just going toīe the second derivative of y with respect to x times this. So times dy dx and then to that, I'm going to add the derivative of this with respect to x. And so, I took the derivative of this, I'm gonna multiply it times that. Now, if I take theĭerivative of negative x with respect to x, that's going to be negative one. I took the derivative of three y squared with respect to y, and took the derivative of y with respect to x. Implicit differentiation, which is really just anĮxtension of the chain rule. Watch the several videos on Khan Academy on As we say, and this is completely unfamiliar to you and I encourage you to The derivative of y with respect to x, so times dy dx. Of three y squared with respect to y, it's going to be six y, and then I have to take Three y squared minus x with respect to x, well, that is going to be, so if I take the derivative First, I'll take theĭerivative of this expression and then multiply that times dy dx, and then I'll take theĭerivative of this expression and multiply it times this. ![]() And so, here I'll apply the product rule. Of the left-hand side with respect to x, and the derivative of the right-hand side with respect to x. And so, let's, let's apply our derivative In fact, if I want, I could, well actually, let me And so now, taking theĭerivative of both sides of this is going to be a littleīit more straightforward. Implicit differentiation one way or the other, so what if I multiply both sides of this times three y squared minus x? Then I get three y squared minus x times dy dx is equal to, is equal to y. Of the right-hand side, I'm gonna have to apply the quotient rule, which is really just comesįrom the product rule, but it gets pretty hairy. That we could try to tackle it, but the way it's written right now, if I take the derivative And now what I wanna do is essentially, take the derivative of both sides again. So dy dy dx is equal to y over three y squared minus x. ![]() Alright, so let's just go to the beginning where they tell us that dy dx is equal to y over three y squared minus x. Where x equals negative one and y is equal to one. The second derivative of y with respect to x squared at the point on the curve ![]()
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