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# Normal to y=𝑒ˣ/x²

## Video transcript

So we have the function f of x is equal to e to the x over x squared. And what I want to do in this video is find the equation, not of the tangent line, but the equation of the normal line, when x is equal to 1. So we care about the equation of the normal line. So I encourage you to pause this video and try this on your own. And if you need a little bit of a hint, the hint I will give you is, is that the slope of a normal line is going to be the negative reciprocal of the slope of the tangent line. If you imagine a curve like this, and we want to find a tangent line at a point, it's going to look something like this. So the tangent line is going to look like this. A normal line is perpendicular to the tangent line. This is the tangent line. The normal line is going to be perpendicular to that. It's going to go just like that. And if this has a slope of m, then this has a slope of the negative reciprocal of m. So negative 1/m. So with that as a little bit of a hint, I encourage you to find the equation of the normal line to this curve, when x equals 1. So let's find the slope of the tangent line. And then we take the negative reciprocal, we can find the slope of the normal line. So to find the slope of the tangent line, we just take the derivative here and evaluate it at x equals 1. So f prime of x, and actually, let me rewrite this a little bit. So f of x is equal to e to the x times x to the negative 2. I like to rewrite it this way, because I always forget the whole quotient rule thing. I like the power rule a lot more. And this allows me to use the power rule. I'm sorry, not the power rule, the product rule. So this allows me to do the product rule instead of the quotient rule. So the derivative of this, f prime of x, is going to be the derivative of e to the x. Which is just e to the x times x to the negative 2, plus e to the x times the derivative of x to the negative 2. Which is negative 2x to the negative 3 power. I just used the power rule right over here. So if I want to evaluate when x is equal to 1, this is going to be equal to-- let me do that in that yellow color like. I like switching colors. This is going to be equal to, let's see, this is going to be e to the first power. Which is just e times 1 to the negative 2, which is just 1 plus e to the first power, which is just e, times negative 2. 1 to the negative 3 is just 1. So e times negative 2. So let me write it this way. So minus 2e. And e minus 2e is just going to be equal to negative e. So this right over here, this is the slope of the tangent line. And so if we want the slope of the normal, we just take the negative reciprocal. So the negative reciprocal of this is going to be, well the reciprocal is 1 over negative e, but we want the negative of that. So it's going to be 1/e. This is going to be the slope of the normal line. And then if we, and our goal isn't just to the slope of the normal line, we want the equation of the normal line. And we know the equation of a line can be represented as y is equal to mx plus b, where m is the slope. So we can say it's going to be y is equal to 1/e-- remember, we're doing the normal line here-- times x plus b. And to solve for b, we just have to recognize that we know a point that this goes through. This goes through the point x equals 1. And when x equals 1, what is y? Well, y is e to the 1st over 1, which is just e. So this goes to the point 1 comma e. So we know that when x is equal to 1, y is equal to e. And now we can just solve for b. So we get e is equal to 1/e plus b. Or we could just subtract 1 over e from both sides, and we would get b is equal to e minus 1/e. And we could obviously right this as e squared minus 1/e if we want to write it like that. But could just leave it just like this. So the equation of the normal line-- so we deserve our drum roll right over here-- is going to be y is equal to 1/e times x, plus b. And b, plus b, is all of this. So plus e minus 1/e. So that right there is our equation of the normal line.