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Tangent to y=𝑒ˣ/(2+x³)

Sal finds the equation of the line tangent to the curve y=eˣ/(2+x³) at the point (1,e/3). Created by Sal Khan.

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  • duskpin ultimate style avatar for user Paige Gladstone
    Could you have used the quotient rule here? Curious because it wasn't used but is in the quotient rule section.
    Would have appreciated more direct uses of the quotient rule here (in these videos) instead of going by way of the product and chain rules.
    (15 votes)
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  • piceratops ultimate style avatar for user Nicolas M
    This video refers to the chain rule, but if you follow 'Calculus 1', the chain rule will only be explained in the next section.
    (11 votes)
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  • aqualine seed style avatar for user as404642
    when tangent line is parallel to y-axis its slope will be?
    (6 votes)
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    • leaf blue style avatar for user Stefen
      The slope of vertical line (a line parallel to the y axis) is undefined. Why? Because slope is defined as change in y divided by change in x. If the line is vertical there is no change in x, that is given that change in x is x(1)-x(0), then x(1)-x(0) = 0 because x(1) = x(2). In that case the calculation for slope results in division by zero, which is undefined.
      (8 votes)
  • blobby green style avatar for user jhomenchak
    starting at around it seems sal forgot to multiply in x^2? im not sure if i missed where he explained it, but i beleive it should have been e/3 + (e(x^2)/3). i appologize if i am mistaken
    (7 votes)
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  • blobby green style avatar for user Shams Sheenwari
    sir
    as we say that the slope of a normal line at a point is the negative reciprocal of tangent line how we can interprate that , does it mean that the slope of normal line is equal to - deltax /delta y ??
    (6 votes)
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    • leafers ultimate style avatar for user KrisSKing
      The normal line is perpendicular to the tangent line. So go back to algebra. Remember two parallel lines have the same slope. Lines that are perpendicular have slopes that are negative reciprocals of each other. So the slope of the normal line is the negative reciprocal of the slope of the tangent line.

      If you are determining -deltax/deltay by finding the derivative of the function (deltay/deltax) and then taking the negative reciprocal, you are fine. But if you try to find -deltax/deltay by taking the negative derivative of the inverse function, that doesn't work.
      (5 votes)
  • blobby green style avatar for user Yomna Ali ElSherif
    why is the derivative of (e power x ) also e power x ?
    (5 votes)
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  • blobby green style avatar for user rigvedj
    What if we want to find the equation of the normal line to the given curve at x = 1 ? as the slope of the tangent line equals 0 at x =1 , the slope of the normal line which is the negative reciprocal of the slope of the tangent line becomes negative infinity. so the usual slope intercept form of the equation of a line does not work for the normal line here. but since the tangent line at x = 1 is parallel to the x-axis, can we say that the normal line at x = 1 will be parallel to the y axis ? and since it passes through the point (1,e/3) , can the equation of the normal line be written as x = 1 ?
    (4 votes)
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    • male robot hal style avatar for user Jesse
      That is correct. You cannot legitimately calculate using the slope of a vertical line, since the slope is infinite. However, you can observe that the normal line is vertical and write the appropriate equation for a vertical line: x=a, or in this case, x=1.
      (4 votes)
  • female robot ada style avatar for user cathalg
    So seeing as this is the Quotient rule section I tried to solve using the quotient rule, here's what I got :
    d/dx e^x/(2+x^3) = d/dx(e^x) (2+x^3) - d/dx(2+x^3) (e^x) / (2+x^3)^2
    = (e^x) (2+x^3) - (3x^2) (e^x) / (2+x^3)^2
    substituting x=1
    = e(2+1) - 3(e) / (2+1)^2
    = 3e -3e / 9
    = 0 / 9
    = 0
    (5 votes)
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  • aqualine seed style avatar for user corrupt11111
    At , Sal uses the chain rule for the derivative of (2+x^3)^-1. Would it not work if he just used the power rule and left it at that?
    (2 votes)
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    • blobby green style avatar for user Creeksider
      No, the power rule applies only when you have x to the n, not when you have some function of x raised to the n. You can see this with an example like (x^2)^3. If you just apply the power rule, you get 3(x^2)^2, but we know that's wrong because (x^2)^3 is x^6, so the answer has to be 6x^5 or something equivalent. You get the right result when you apply the chain rule.
      (6 votes)
  • leaf blue style avatar for user Jan Staněk
    What about the topic "Tangents to polar curves", is there gonna be some video guide sometimes?
    I tried the exercise, but it's confusing. I'm writing it here, cuz there's no video in there to comment below.
    (1 vote)
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Video transcript

We have the curve y is equal to e to the x over 2 plus x to the third power. And what we want to do is find the equation of the tangent line to this curve at the point x equals 1. And when x is equal to 1, y is going to be equal to e over 3. It's going to be e over 3. So let's try to figure out the equation of the tangent line to this curve at this point. And I encourage you to pause this video and try this on your own first. Well, the slope of the tangent line at this point is the same thing as the derivative at this point. So let's try to find the derivative of this or evaluate the derivative of this function right over here at this point. So to do that, first I'm going to rewrite it. You could use the quotient rule if you like, but I always forget the quotient rule. The product rule is much easier for me to remember. So I can rewrite this as y is equal to-- and I might as well color code it-- is equal to e to the x times 2 plus x to the third to the negative 1 power. And so the derivative of this, so let me write it here. So y prime is going to be equal to the derivative of this part of it, e to the x. So the derivative of e to the x is just e to the x. Just let me write that. So we're going to take the derivative of it. And that's what's amazing about e to the x, is that the derivative of e to the x is just e to the x times this thing. So times 2 plus x to the third to the negative 1. And then to that we're going to add this thing. So not its derivative anymore. We're just going to add e to the x times the derivative of this thing right over here. So we're going to take the derivative. So we can do the chain rule. It's going to be the derivative of 2 plus x to the third to the negative 1 power with respect to 2 plus x to the third times the derivative of 2 plus x to the third with respect to x. So this is going to be equal to negative-- I'll write it this way-- negative 2 plus x to the third to the negative 2 power. And then we're going to multiply that times the derivative of 2 plus x to the third with respect to x. Well, derivative of this with respect to x is just 3x squared. And of course, we could simplify this a little bit if we like. But the whole point of this is to actually find the value of the derivative at this point. So let's evaluate. Let's evaluate y prime when x is equal to 1. Y prime of 1 when x is equal to 1. This thing will simplify to-- let's see, this is going to be e times 2 plus 1 to the negative power. So that's just going to be 1/3, right? 2 plus 1 to the negative 1. So that's 3 to the negative 1. That's 1/3. So that's times 1/3 plus e to the first power. Now let's see, what does this do? This part right over here, this is 2 plus 1 to the negative 2 power. So this-- actually, let me-- I don't want to-- so this part right over here is going to be, let's see, this is going to be-- I don't want to make a careless mistake here-- is 3 to the negative 2 power. So 3 squared is 9. 3 to the negative second power is 1/9. And so it's going to be 1/9. Well, you're going to multiply this negative there. So it's negative 1/9. And then we're going to multiply that times 3 times 1. So it's negative 1/9 times 3. Times 3 right over here. So it's negative 3/9 or negative 1/3. So times negative 1/3. And all I did here is I substituted 1 for x and evaluated it. Now this is interesting. I have essentially-- let me rewrite this. This is equal to e over 3 minus e over 3, which is equal to 0. So the slope of the derivative when x is equal to 1 is 0, or the slope of the tangent line is equal to 0. This simplified to a pretty straightforward situation. If I wanted to write a line in slope intercept form, I could write it like this. y is equal to mx plus b, where m is the slope and b is the y-intercept. Now we know that the slope of the tangent line at this point, it has a slope, is 0. So this is going to be 0. So this whole term is going to be 0. So it's just going to have the form y is equal to b. This is just going to be a horizontal line. So what is a horizontal line that contains this point right over here? Well, it contains the value y is equal to e over 3. So this is a horizontal line. It has the same y value the entire time. So if it has the y value e over 3, then we know the equation of the tangent line to this curve at this point is going to be y is equal to e over 3. Another way you could think about this right over here is, well, let's substitute when x is equal to 1. Well, there's not even an x here. But when x is any value, y is equal to e over 3, you get b is equal to e over 3, or you'd get y is equal to e over 3. So it's just a horizontal line. So let's actually visualize this, just to make sure that this actually makes sense. So let me get my graphing calculator out. And so I'm in graphing mode. If you wanted to know how to get there, you literally can just go to graph, y equals, and I will do-- so e to the x power divided by 2 plus x to the third power. That looks right. And I actually set the range ahead of time to save time. So let me graph this. So let's see. Ooh, it does all sorts of interesting things. All right. Oh, look at that. All right, so now we can trace to get to when x is equal to 1. x equals 1. Right over there, you see y is equal to e over 3, which this is kind of its decimal expansion right over here. And it does look like the slope right over here is 0, that the tangent line is just going to be a horizontal line at that point. So that makes me feel pretty good about our answer.