In Exercises 21–48, find the derivative of y with respect to t...

Question

In Exercises 21–48, find the derivative of y with respect to the appropriate variable.

45. y=cos(x-arccos(x))

Accepted Answer

Hello there. Today we are going to 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 DY by DX given that Y is equal to sin of x plus inverse cosine of X. OK, so it appears for this particular problem, we're ultimately asked to solve for Dy by DX given that we are provided a specific equation for the equation of Y. And using all the information that's provided to us, we're ultimately, once again, trying to solve for DY by DX. So now that we know what we're ultimately trying to solve for. Our first step that we need to take is we need to set the inner and outer functions in order to recall and apply the chain rule. So let us state that U of X, our function U X will be equal to X plus inverse cosine of X. And let us state that. Y of you will be equal to sin of you. Fantastic. And when we recall and use the chain rule, it will give us that DY by DX will be equal to DY by DU multiplied by DU by DX. Fantastic. So our second step now is we need to compute the outer derivative DY by DU. Sense Y is equal to sin of u. So that will mean, based on this information that DY. By DU. Will be simply equal to. Co-sign of you. Fantastic. So now, moving right along here, our third step now is we need to compute the inner derivative. We need to compute DU by DX, and we need to note that as u of X is equal to X plus inverse cosine of X, that will mean that we can go ahead. And write that DU by DX will be equal to D by DX X plus D by DX of. Inverse cosine. Of X Which will be equal to 1 + D by Dx of inverse cosine of X. Fantastic. So, we now need to recall and use the standard derivative method. So we need to take D by DX of inverse cosine of X, which will be equal to -1 divided by the square root of 1 minus x 2. Multiplied by parentheses of -1 is less than x and x is less than 1. So, that will mean that undoubtedly that DU by DX, so DU by DX will be simply equal to 1 minus 1 divided by the square root of 1 minus x to the power of 2. Fantastic. So, for our 4th and final step is we need to combine our worst. Once again, we're trying to solve for Dy by DX and in order to do that, we need to combine our results, and we need to make sure that the chain rule is applied explicitly. So when we do that, we will find that DY by DX is equal to cosine of u. Multiplied by parentheses 1 minus. So it's gonna be 1 minus 1 divided by the square root of 1 minus X2. Which will be equal to cosine of x plus inverse cosine of x multiplied by 1 minus 1 divided by the square root of 1 minus x 2 in parentheses. Which will be equal to cosine of x plus inverse cosine of x multiplied by 1 minus the square root of 1 minus x 2 divided by 1 minus x 2. Which will be all equal to. When we simplify cosine of. X plus inverse cosine of X. Multiplied by 1 minus x to the power of 2. Minus the square root of 1 minus x to the power of 2. All divided by 1 minus x of 2. And don't forget your parenthesis, so that will mean that our final answer without a doubt will be DY by DX is equal to cosine of x plus inverse cosine of x multiplied by parentheses of 1 minus x to the power of 2. Minus the square root of 1 minus x to the power of 2. Divided by 1 minus x to the power of 2. And this is our final answer. Hooray, we did it. So going back up to the top to recap what we've learned and to look at the problem itself, we can safely say that without a doubt, our final answer for Dy by DX is once again as follows. So Dy by dx is equal to cosine of x plus the inverse cosine of x multiplied by parentheses of 1 minus x 2 minus the square root of 1 minus x 2, all divided by 1 minus x 2. 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 21–48, find the derivative of y with respect to the appropriate variable.45. y=cos(x-arccos(x))

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In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
45. y=cos(x-arccos(x))