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.

41. y= x arcsin(x) + √(1-x²)

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 parenthesis of 1 minus x2 multiplied by inverse of X plus x multiplied by the square root of 1 minus X to the power of 2. OK. So it appears for this particular problem we're asked to ultimately solve for DY by DX given that we are given a specific equation for Y by the problem itself. 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 differentiate the first term, which is parentheses 1 minus X to the power of 2 multiplied by inverse of X by the product rule. Which means we need to let, as we should recall, we need to let UX be equal to 1 minus X to the power of 2, and we need to let V of X be equal to inverse of X, then that will mean we can go ahead and write that U. X is equal to -2 X and that V of X will be equal to 1 divided by the square root of 1 minus X to the power of 2. Therefore, we could write that D by DX of parentheses. So once again, D by DX of parentheses. Of 1 minus x to the power of 2 multiplied by inverse of X, which will be equal to U X multiplied by V X plus UX multiplied by V. of X, which will be equal to -2 multiplied by x multiplied by inverse of x + 1 minus x 2 multiplied by 1 divided by the square root of 1 minus X to the power of 2. OK, so we need to note that, let's make a note. We need to note that parentheses 1 minus X to the power of 2 divided by the square root of 1 minus X 2 will be equal to the square root of 1 minus X to the power of 2. So, as a result, we can go ahead and write that D by DX of parenthesis of 1 minus X to the power 2 multiplied by inverse sin of X will be. Subtracted from 2 multiplied by X, multiplied by inverse sign of X plus the square root of 1 minus X to the power of 2. Hooray, we are on a roll. So our second step now, so this step number 2, is we need to differentiate the second term, which our second term is X multiplied by the square root of 1 minus X to the power of 2, and we need to differentiate this second term by recalling and using the product rule. So let us state that P of X is equal to X and that Q. X is equal to the square root of 1 minus x to the power of 2. So that will mean then that P X will be equal to 1 and that Q X will be equal to 1 divided by 2 multiplied by the square root of 1 minus X to the power of 2. Multiplied by -2 X, which will be equal to negative X divided by the square root of 1 minus X to the power of 2. Fantastic. So, that will mean that D so one thing that will mean that D by DX. Of X multiplied by the square root of 1 minus X to the power of 2 will be equal to PX multiplied by Q X plus PX multiplied by Q X. Which will be equal to the square root of 1 minus X2 minus X 2 divided by the square root of 1 minus X2. Which will be all equal to 1 minus 2 multiplied by X to the power of 2, all divided by the square root of 1 minus X to the power of 2. Fantastic. So now, our step number 3, so this will be step number 3, is that we now need to add the derivatives from steps 2 and 3. So we can go ahead and write DY by. D X is equal to -2 multiplied by X multiplied by inverse of X plus the square root of 1 minus X to the power of 2, plus 1 minus 2 multiplied by x 2 all divided by the square root of 1 minus X2. And once again, what we are doing is we're adding the derivatives from steps 2 and 3. Fantastic. So now, we need to combine like terms with square root of 1 minus X to the power of 2. By writing, so by writing that the square root of 1 minus. X 2 will be equal to parentheses 1 minus x 2 divided by the square root of 1 minus X to the power of 2. So we can go ahead and write that the square root of 1 minus X2 plus 1 minus 2 multiplied by x 2 all divided by the square root of 1 minus X 2 will be equal to 1 minus X2 + 1 minus 2 x 2. So, B minus 2 multiplied by X to the power 2. All divided by the square root of 1 minus X2, and when we simplify this expression to its most simple form, we will find that it's equal to 2 minus 3 multiplied by x to the power of 2, all divided by the square root of 1 minus X to the power of 2. So, therefore, we can go ahead and write that DY by DX will be equal to -2 multiplied by X multiplied by inverse of X plus 2 minus 3 multiplied by X to the power of 2, all divided by 1 minus X to the power of 2. Noting that the absolute value of X is less than 1, and this is our final answer for DY by DX. Hooray, we did it. We solved for our final answer. So going back to look at the problem itself, to recap what we've learned and to state our final answer one last time, we will find that our final answer once again. Without a doubt, will be as follows. So our answer for Dy by DX will be, so once again, Dy by D X is equal to -2 multiplied by x multiplied by inverse of X plus 2 minus 3 multiplied by x 2 all divided by the square root of 1 minus X2. 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.41. y= x arcsin(x) + √(1-x²)

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