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the differentiable functions x and y are related by the following equation the sine of X plus cosine of Y is equal to square root of two they also tell us that the derivative of X with respect to T is equal to five they also ask us find the derivative of Y with respect to T when y is equal to PI over 4 and 0 is less than X is less than PI over 2 so given that they are telling us the derivative of X with respect to T and we want to find the derivative of Y with respect to T it's a safe assumption that both x and y are functions of T so you could even rewrite this equation right over here you could rewrite it as sine of X which is a function of T plus cosine of Y which is a function of T is equal to square root of 2 now it might confuse you a little bit you're not used to seeing X as a function of a third variable or Y is a function of something other than X but remember x and y are just variables this could be f of T and this could be G of T instead of X of T and yfd and that might feel a little bit more natural to you but needless to say if we want to find dy DT what we want to do is take the derivative with respect to T of both sides of this equation so let's do that so we're going to do it on the left hand side so it's going to be retake that with respect to T derivative of that with respect to T we're gonna take the derivative of that with respect to T and then we're going to take the derivative of the right hand side this constant with respect to T so let's think about each of these things so what is this let me do this in a new color the stuff that I'm doing in this aqua color right over here how could I write that so I'm taking the derivative with respect to T I have sine of something which is itself a function of T so I would just apply the chain rule here I'm first going to take the derivative with respect to X of nine of X I could write sine of X of T but I'll just revert back to the sine of X here for simplicity and then I would then multiply that times the derivative of the inside you could say with respect to T times the derivative of X with respect to T this might be a little bit counterintuitive to how you've applied the chain rule before when we only dealt with X's and Y's but all that's happening I'm taking the derivative of the outside of the sine of something with respect to that something in this case it is X and then I'm taking the derivative of the something in this case X with respect to T well we can do the same thing here for this second term here so I want to take the derivative with respect to Y of I guess you could say the outside of cosine of Y and then I would multiply that times the derivative of Y with respect to T and then all of that is going to be equal to what well the derivative with respect to T of a constant square root of two is a constant it's not going to change as T changes so it's derivative it's rate of change is zero all right so now we just have to figure out all of these things so first of all the derivative with respect to X of sine of X is cosine of X times the derivative of X with respect to T I'll just write that out here derivative of X with respect to T and then we're going to have it's going to be a plus here the derivative of Y with respect to T so plus the derivative of Y with respect to T I'm just swapping the order here so that this goes out front now what's the derivative of cosine of Y with respect to Y well that is negative sine of Y and so actually let me just put the sine of Y here and then I want to have a negative so let me erase this and put a negative there and that is all going to be equal to zero and so what can we figure out now they've told us to the derivative of X with respect to T is equal to 5 they tell us that right over here so this is equal to five we want to find the derivative of Y with respect to T they tell us what Y is y is PI over four this Y is PI over four so we know this is PI over four and let's see we have to figure out what we still have two unknowns here we have we don't know what X is and we don't know what the derivative of Y with respect to T is this is what we need to figure out so what would X be what would X be when y is PI over four well to figure that out we can go back to this original equation right over here so when y is PI over four you get let me write down sine of X plus cosine of PI over four is equal to square root of two cosine of PI over four we revert to our units or we think about our unit circle we're in the first quadrant if you think in degrees that's a 45 degree angle that's going to be square root of two over two and so we can subtract square root of two over two from both sides which is going to give us sine of X is equal to well if you take square root of two over two from square root of two you're taking half of it away so you're going to have half of it less so square root of two over two and so what x value when I take the sine of it and remember where the angle if we're thing when the unit circle is going to be in that first quadrant X is an angle in this case right over here well that's going to be once again PI over 4 so this tells us that X is equal to PI over four when y is equal to PI over four and so we know that this is PI over four as well so let me just rewrite this because it's getting a little bit messy so we know that five times cosine of PI over four minus dy DT the derivative of Y with respect to T which is what we want to figure out times sine of PI over four is equal to zero is equal to zero and let me put some parentheses here just to clarify things a little bit all right so let's see now it's just a little bit of algebra cosine of PI over four we already know is square root of 2 over 2 sine of PI over 4 is also square root of 2 over 2 now let's see what if we divide both sides of this equation by square root of 2 over 2 well what's that going to give us well then this square root of 2 over 2 divided by square root of 2 over square root of 2 over 2 divided by square root of 2 over 2 is going to be 1 square root of 2 over 2 divided by square root of 2 over 2 is going to be 1 and then 0 divided by square root of 2 over 2 is just still going to be 0 and so this whole thing simplifies to 5 times 1 which is just 5 minus the derivative of Y with respect to T is equal to 0 and so there you have it you add the derivative of Y with respect to T to both sides and we get the derivative of Y with respect to T is equal to 5 when all of these other things are true when the derivative of X with respect to T is 5 and the derivative and why I should say is equal to PI over 4

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