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Current time:0:00Total duration:7:17

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 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 and the value 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 x times 2 plus X to the 3rd 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 this this part of it e to the X so the derivative of e to the X is just e to the X so 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 3rd 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 times the derivative of this thing right over here so we're going to apply 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 right this way negative two plus X to the third to the negative two power to the negative two power and then we're going to multiply that times the derivative of two plus X to the third with respect to X well the derivative of this with respect to X is just 3x squared 3x 3x squared and of course we could we could simplify this a little bit 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 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 e times 2 plus 1 to the negative 1 power so that's just going to be 1/3 right 2 plus 1 to the negative 1 so it's 3 to the negative 1 that's 1/3 so let me just times 1 3 plus E is to the first power 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 don't want to I don't want to so this part right over here is going to be let's see this is going to be or make a careless mistake here this is 3 to the negative 2 power so 3 squared is 9 3 to the negative 2nd 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 negative 3 ninths or negative 1/3 so x 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 three minus e over three which is equal to zero so the slope of the the derivative when X is equal to one is zero or the slope of the tangent line is equal to zero and so this is actually pretty this is a pretty this simplified to a pretty straightforward situation if I wanted to write 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 zero so this is going to be zero so this whole term is going to be zero 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 three so this is a horizontal line it has the same y value of the entire time so if it has about Y value of e over three then we know that the equation of the tangent line to this curve at this point is going to be y is equal to e over is equal to e over three another way you could think about this right over here is well let's substitute what X is equal to one well there's not even an X here but when X is when X is any value Y is equal to e over three you get B is equal to e over three or you'd get Y is equal to e over three so it's just a horizontal line so let's actually visualize this just to make sure that this actually makes sense let me get my graphing calculator out get my graphing calculator out and so I'm in graphing mode if you wanted to if you wanted to know how to get there you literally you literally can just go to graph y equals and I will do so e to the X power divided by two plus X to the third power X to the third power that looks right and I you set the range ahead of time to save time so let me graph this so let's see what what 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