In Exercises 41–58, find dy/dt.

y = √(3t + (√...

Question

In Exercises 41–58, find dy/dt.

y = √(3t + (√2 + √(1 − t)))

Accepted Answer

Welcome back, everyone. In this problem, we want to find the derivative of Y equals the square root of 4 X plus the square root of 3 plus the square root of 2 minus X. Now, before we solve for the derivative of Y, there are a few things that we can know, OK? Now, first of all. If we rewrite or if we take a good look at our function here, we can tell that this is a composite function. And because of that, that means we'll need to use the chain rule and at some point also the power rule of differentiation. Recall The the power rule tells us that if we differentiate a term, OK, X to the power of N, it's going to be equal to N multiplied by X to the power of N minus 1, OK? In other words, we bring down the power and then subtract 1 from the power in the term. And that also applies even if we were multiplying this uh term by a multiple, say a multiple C. Next, we also know that for the chain rule it basically tells us that the derivative of a composite function F of G of X is going to be equal to the derivative of F of gefX multiplied by the derivative of G of X. In other words, we differentiate the autoer function and multiply it by the derivative of the inner function. And know, notice here that we'll be working with roots a lot, OK. So it would be good for us to make a note here of the derivative of a term for a root. In this case, the derivative of root X, which is going to be equal to 1 divided by 2 root X. So now with all this information, we can go ahead and try to differentiate why. Now when we do that. As we said, since Y is a composite function, then first we'll differentiate the alter function, OK, which we know the square root, uh, is gonna be basically a fraction of pole raise to the power of a half. So we bring down our po half multiplied by 1 divided by, OK, 1 divided by Root square root of 4 X plus the square root of 3 plus the square root of 2 minus X, OK. Multiplied by the derivative of our inner function here. In this case, it's gonna be 4 X plus route 3 + route 2 minus X. No If we're going to solve, it would be good for us to try to find the derivative of our inner function. So let's go ahead and do that. Let's differentiate, OK? So now, Finding the derivative with respect to X. 4 X plus root 3 + route 2 minus X, OK. Then we know that this is gonna be equal to 4 X 1 minus 1. We subtract one from the poor. We apply our poor rule here plus the derivative of this term, Route 3 plus route 2 minus X, and that, again, if we use the same idea as before, it's gonna be 1 divided by 2 route 3 plus route 2 minus X, OK. Multiplied by the derivative of the inner function, which in this case is now 3 + root 2 minus X, OK? Now if we go ahead here. Then that's gonna become 4 + 1 divided by 2 root 3 plus root 2 minus X. Multiplied by. 0 + 1 divided by 2. Route 2 minus X multiplied by the derivative of 2 minus X again we apply our chain rule. Now if I make some space and come over to the right, then we just need to find our derivative here of uh 2 minus X to simplify. So now, this is gonna give us 4 plus. One divided by. Because we know we have here 4 plus, and let me just make some space here cause we originally had 41 divided by 2, Route 3 plus route 2 minus X. Multiplied by 1 divided by 2 root 2 minus x multiplied by -1, the derivative of 2 minus X. No, we can start to simplify. If we multiply through this term by -1, then this is gonna become 4 minus 1. if, and, and we combine them all divided by 4 multiplied by root 3 plus root 2 minus X multiplied by root 2 minus X. And now, if we combine both of these terms, OK, then we know it's gonna be 4 divided by 1 minus the term that we have here, which, when we simplify, no, that's gonna be equal to 16 multiplied by route 3 plus route 2 minus X. OK. We're, we're uh multiplying, we're combining both fractions here. OK. Multiplied by route 2 minus XK minus 1. Now this is all divided by 4 root 3 plus root 2 minus X multiplied by root 2 minus X. So this is gonna be the derivative of our inner function. Now that we have that, we can go back to the derivative of Y. Because no, this means then that the derivative of Y, we write back the term we had before is gonna be 1 divided by 2. Route 4 X plus route 3 + route 2 minus X, OK. Multiplied by this derivative of the inner function, which, uh, in the interest of time, I'll just copy and paste here, OK. And now if we go through and multiply, if we multiply through for our terms. Then we should find that we end up with 16 multiplied by, well, it's 1 multiplied by this term, so we'll basically just have that term in our denominator, OK? In our, in our numerator, sorry. -1 And all of this is being divided by 2 multiplied by 48, multiplied by the root of the denominate or multiplied by that term in our first expression here. So that's root 4 X plus root 3 + root 2 minus X, OK. Multiplied by route 3 plus route 2 minus X multiplied by route 2 minus X. And that this is going to be the derivative of Y. Thanks a lot for watching everyone. I hope this video helped.

In Exercises 41–58, find dy/dt.

y = √(3t + (√2 + √(1 − t)))

Key Concepts

Chain Rule

Derivative of Square Root Function

Implicit Differentiation

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