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The equation of the tangent line to a curve at the point (2,4) is $y=3 x-2$If $\frac{d^{2} y}{d x^{2}}=6 x-7$ at any point on the curve, find its equation.

$$y=x^{3}-\frac{7}{2} x^{2}+5 x$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 2

Applications of Antidifferentiation

Integrals

Missouri State University

Campbell University

Oregon State University

University of Nottingham

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

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find an equation of the ta…

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Find an equation of the ta…

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Find the equation of the t…

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whether in this problem, we're going to find an equation for the tangent line. But the curve e to the two x minus y equals X squared over. Why? And we're going to do this at the point two comma for Okay, So we're going to use point slope form at the end, and we're just going to need this slope since we have the point and to find slope. As always, we're going to want the derivative. Okay, um uh, we're going out these quotient rule on the right side, which is all right. If we multiply through my ex, why? Sorry? We would need a We need a product rule on the left. So let's But let's not be fancy here. We'll just take the derivative as it is. Either way, it'll be a little bit of work. That's okay. On the left side, the derivative of E to the two x minus Y is you know how these work The Druid of E to any function of X, it's just itself. And now we need to use the chain rule I need multiplied by the derivative of this blue exponents. There was the derivative of two X minus y. Well, the derivative of to access to the derivative of why is just one. But then, as always, we need to multiply by white prime any time to take the derivative of why for these implicit problems Um good. So the right side now the root of X squared over while need quotient rule. So we get the denominator times the derivative, the numerator minus numerator times the derivative of the denominator derivative of y as before his wife prime and all over the denominator squared. That's just the quotient rule for the right side. Okay, um, let's do a couple things at the end. We're going to have to isolate the wide primes. So we have them all over the place here on the left. Why don't we expand? We'll just multiply this. So that will give us two years to x minus Y minus. Why, prime times Eat of the two x minus y and the rights again. We're just going thio clean this up just a little bit, okay? And finally get rid of the fractions. Why don't we multiply everything by y squared? So that means the right side will simply be the numerator and so let's make sure this is that was supposed to be a dot here, so we really let's not confuse ourselves. Let's make sure it's right. That is X squared Y prime. It's just multiplication. That's all we have on the right side. On the left. We're gonna have why squared times, everything. They're so weaken. Do the math. There's get to y squared E to the two x minus. Y minus. Be careful. Otherwise, because things are gonna look the same. It's why squared times why prime times eating two X minus Y Okay, let's look to see where are y prime terms are. There's one over here and there's one over here, so there's one on each side that has a Y prime in it. We're going to have to move things around, and we're going to subtract his blue term from the left from both sides so that it ends up on the right. So after we do a little algebra, that's just make sure that I'm left. What are we gonna have? We're gonna keep the red term, and we're also gonna and that green term from the right, that's all we're gonna have on the right side. We're gonna keep our two x y. We also had a track that blue term from the left, minus two my square D to the two x minus Y Okay, this way. All the white prime terms air together on the left. And now we can factor out the wide prime. It'll give us negative. Why squared e to the two X minus y plus X squared on the right. We have exactly what we had before, and we are basically done with finding the derivative. Just divide both sides by the parentheses, and we have our derivatives. So now that we have the derivative finish reading it out here, now that we have the derivative, we just need to plug in the actual point so that we get a slope. So is create a little space for ourselves here. We need our slope specifically at the point to come a four. That means we're just going to plug into for X four for wine and be very careful with her math and see what we get. So on top, we're just looking at our derivative from the bottom here to see what we do two times X is too wise for my ass two times. Why is four e to the power C. Two times two minus for okay, we'll clean this all up later. Let's just plug things in for now. Minus four. Squared E to two times X minus wives for caused. Two squared. OK, now just a lot of arithmetic. So in the top two times two is four times for this. 16. Concede to times four Squared is 32 e to the Let's hear the exponents of E. There's two times two minus four and luckily for us is zero, and eat of the zero is just one. And on the bottom it's the negative. Four square. It's negative. 16 e Once again, this exponents here for E comes out to zero. So you just get one plus to score. It is for okay and so on. Top 16 minus 32 is negative. 16 on bottom Negative. 16 plus four Negative. 12. Who can reduce that a little bit. We should get positive 4/3. Okay. And therefore we're ready for the equation of our tangent line. We can use the point slope form. So the Y coordinate was for slope we just got was 4/3 and the x coordinates where the point was to. So this here is an equation for the tangent line. That's all they asked us for so we could be done. If you want to put it into point slope form. I'm sorry. If you want to put it into slope intercept form, you certainly can. You know how to do that. Just rearrange things. But this answer is certainly ballot. Hopefully, they don't.

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