In Exercises 41–58, find dy/dt.

y = 4 sin(√(1...

Question

In Exercises 41–58, find dy/dt.

y = 4 sin(√(1 + √t))

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So, first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Find the derivative of Y equals 8 multiplied by sine of the square root of 4 plus the square root of X. Awesome. So it appears for this particular problem, we're ultimately trying to figure out what the derivative of this equation of Y is. So now that we know that we're trying to find the derivative of Y. First off, we need to recall that D by DX sine of X is equal to cosine of X. We also need to recall and note that D by DX of the square root of X will be equal to 1 divided by 2 multiplied by the square root of X. We also need to recall and use the following rule. For differentiation. So let us recall the constant rule which states that D by DX of some constant value is going to be equal to 0. We also need to recall and use the chain rule which states that D by DX of F of G of X is going to be equal to F of G of X multiplied by G of X. Fantastic. So now we can officially start to solve for Y where this prime sample indicates the first derivative. So the first derivative of Y is what we're ultimately trying to solve for. So Y is going to be equal to D by DX of. 8 multiplied by sine of the square root of 4 plus the square root of X. Fantastic. So now, What we need to do is we need to take Y is equal to, since we're now what we're trying to do is we're trying to perform our derivatives and simplify as we go. So when we perform our first simplification and start to perform our derivatives, we will find that Y is equal to 8 multiple. by cosine of the square root of 4 plus the square root of X, multiplied by D by DX of the square root of 4 plus the square root of X. Fantastic. Moving right along here. So now we could take D by DX of the square root of 4 plus the square root. X is equal to 1 divided by 2 multiplied by the square root of 4 plus the square root of X, multiplied by D by DX of 4 plus the square root of X. So then we can go ahead and write that D by DX of the square root of 4 plus the square root of X will be equal to. Drum roll, 1 divided by 2 multiplied by the square root of 4 plus the square root of X multiplied by 0 + 1 divided by 2 multiplied by the square root of X. Fantastic, moving right along here. Then we could go ahead and write that D by DX of the square root of 4 plus the square root of X. Will be equal to 1 divided by 2 multiplied by the square root of 4 plus the square root of X, multiplied by 1 divided by 2 multiplied by the square root of X. Fantastic. So now, we can go ahead and write that D by D X of the square root of 4 plus the square root of X will be equal to when we simplify one last time, will be 1 divided by 4 multiplied by the square root of 4. X plus x multiplied by the square root of X. So to quickly recap here, when we perform a derivative for D by DX of the square root of 4 plus the square root of X, we will find that it's equal to 1 divided by 4 multiplied by the square root of 4 X plus x multiplied by the square root of X. So that will mean when we solve for Y, we will find that Y will be equal to 8 multiplied by cosine of the square root of 4 plus the square root of X multiplied by 1 divided by 4 multiplied by the square root of 4 X plus X multiplied by the square root of X. Fantastic. So, that will mean when we simplify. So now, when we simplify one last time, we will find that our final answer, without a doubt for Y will have to be, so Y is equal to 2 multiplied by cosine of the square root of 4 min, so it's gonna be 2 multiplied by cosine of square root of 4 plus the square root of X. Awesome. So once again, just to quickly note is we're taking our expression that we saw that we found for Y previously and we're simplifying it one last time. So ultimately when we perform any derivatives, we want it to be in its most simple form, so where we can't simplify it anymore. So when we put it in its most simple form, we will find that Y is equal to 2 multiplied by cosine of the square root of 4 plus the square root of X, all divided by. The square root of 4 X plus x multiplied by the square root of X, and that's it. That is our final answer. Hooray, we did it. So going back to look at our prom itself to quickly recap what we've learned together, our final answer once again without a doubt has to be. So once again, our final answer is Y is equal to 2 multiplied by cosine of the square root of 4 plus the square root of X, all divided by the square root of 4 X plus X multiplied by the square root of X. And that's it. We've officially solved for this problem. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

In Exercises 41–58, find dy/dt.

y = 4 sin(√(1 + √t))

Key Concepts

Chain Rule

Derivative of Sine Function

Derivative of Square Root Function

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