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Current time:0:00Total duration:7:11

AP.CALC:

FUN‑3 (EU)

, FUN‑3.C (LO)

, FUN‑3.C.1 (EK)

so let's say that we have y is equal to the secant of 3 PI over 2 minus X and what we want to do is we want to figure out what dy/dx is the derivative of Y with respect to X is at X equal PI over 4 like always pause this video and see if you could figure it out well as you can see here we have a composite function we're taking the secant not just of X but you could view this as of another expression that I guess you could define or is of another function so for example if we call this right over here U of X so let's do that so if we say U of X is equal to 3 PI over 2 minus X and we could also figure out u prime of X is going to be equal to derivative of 3 PI over 2 that's just going to be 0 derivative of minus X well that's just going to be minus 1 and you could just view that as a power rule it's 1 times negative 1 times X to the 0 power which is just 1 so there you go so we could view this as the derivative of secant with respect to U of X and when we take the derivative the derivative of secant with respect to U of x times the derivative of U with respect to X and you might say well what about the derivative of secant well in other videos we actually prove it out and you could actually read arrive it secant is just 1 over cosine of X so it comes straight out of the chain rule so in other videos we proved that the the derivative of the secant of X of secant of X is equal to is equal to sine of X over cosine of X over cosine of x squared so if we're trying to find the derivative of Y with respect to X well it's going to be the derivative of secant with respect to U of x times the derivative of U with respect to X so let's do that the derivative of secant with respect to U of X well instead of seeing an X everywhere you're going to see a U of X every so this is going to be sine of U of X sine of U of X and I could I don't have to write U of X I could write 3 PI over 2 minus X but I'll write U of X right over here just to really visualize what we're doing so sine of U of x over over cosine squared of U of X cosine squared let me do those parentheses in the blue color just to make sure that you identify it with a trig function so cosine squared of U of X U of X so that's the derivative of secant with respect to U of X and then the chain rule tells it's going to be that times u prime u prime of X so what is this going to be equal to well I could just substitute back this is going to be equal to I will write it like this sine of U of X which is 3 PI over 2 minus X and I'll fill that in a second over cosine of U of x squared times u prime of X U of X is 3 PI over 2 minus X 3 PI over 2 minus X and then u prime of X we all we already figured out is negative 1 so I could write x negative 1 oh yeah let me just leave it out there for now I could I could have just put a negative out front but I really want you to be able to see what I'm doing here and now we want to evaluate at x equals PI over 4 so when that is equal to PI over 4 PI over 4 so let's see this is going to be this is going to be equal to sine of what's 3 PI over 2 minus PI over 4 I'll do that over here so if you have a common denominator that is 6 PI over 4 that's the same thing as 3 PI over 2 minus PI over 4 sorry minus PI over 4 is equal to 5 5 PI over 4 so it's sine of 5 PI over 4 five PI over four over cosine squared of five PI over four and then times negative one I could just put that out here now what is sine of five PI over four and cosine squared of five PI over four well I don't have that memorized but let's actually draw a unit circle and we should be able to we should be able to figure out what that is so a unit circle I try to hand draw it as best as I can please forgive me that this circle does not look really like a circle all right okay so let me just remember my angles so I in my brain I sometimes convert into degrees PI over four is 45 degrees this is PI over 2 this is 3 PI over 4 this is 4 PI over 4 this is 5 PI over 4 lens you right over there so if if you wanted to see where the you intersect the unit circle this is at the point this is at the point your x-coordinate is negative square root of 2 over 2 negative square root of 2 over 2 and your y-coordinate is negative square root of 2 over 2 if you're wondering how I got that I encourage you to review the unit circle and some of the standard angles around the unit circle you'll see that in the trigonometry section of Khan Academy but this is enough for us because the sine is the y-coordinate it's the y-coordinate here so negative square root of 2 over 2 so this is negative square root of 2 over 2 and then the cosine is the x-coordinate which is also negative square root of 2 over 2 what's going to be that squared negative square root of 2 over 2 we're squaring it so if we square this it's going to be calm this is going to become it's going to become positive and then square root of 2 squared is 2 and then 2 squared is 4 so it's 1/2 so this is the denominator is equal to 1/2 C in the numerator this negative cancels that with that negative and so we are left with we deserve a little bit of a drumroll that we are left with square root of two over two that's the numerator divided by one-half well that's the same thing as multiplying by two so we are left with positive square root of two is the slope of the tangent line to the graph of y is equal to this when X is equal to PI over four pretty exciting

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