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### Course: Calculus, all content (2017 edition) > Unit 2

Lesson 22: Quotient rule# 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.

## Want to join the conversation?

- This video refers to the chain rule, but if you follow 'Calculus 1', the chain rule will only be explained in the next section.(36 votes)
- That is fine.

You can do this perfectly using just the quotient rule.(0 votes)

- 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.(23 votes)- i used the quotient rule and got the same asnwer(3 votes)

- when tangent line is parallel to y-axis its slope will be?(6 votes)
- 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.(10 votes)

- starting at around4:10it 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)
- No, if you put in x=1 you get x^2=1.

So he just multiplied by 1.(3 votes)

- 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)- 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)

- why is the derivative of (e power x ) also e power x ?(5 votes)
- After understanding the basics of differentiation,

Watch this video first:

https://www.khanacademy.org/math/calculus/differential-calculus/der_common_functions/v/proof--d-dx-ln-x----1-x

and then this video:

https://www.khanacademy.org/math/calculus/differential-calculus/der_common_functions/v/proof--d-dx-e-x----e-x(5 votes)

- 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)
- 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)

- 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) - At2:25, 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)
- 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)

- 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.(2 votes)

## 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.