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.

37. y=s√(1-s²) + arccos(s)

Accepted Answer

Welcome back everyone. Find dyy divided by dx, given that y equals x multiplied by the square root of 1 minus x2 plus inverse of sin of x. So for this problem, we want to find the DY divided by DX and we have a sum. So we can first of all differentiate. X square root of 1 minus x2 with respect to x. And then we can add the derivative of inverse of sin of x. With respect to X. So, for the first derivative, let's notice that we have a product of two functions. So we can basically apply the product rule, and according to the product rule, we take the derivative of the first function, so that'd be the derivative of X or basically Dx divided by DX. Multiplied by the 2nd function which is square root of 1 minus x 2. Plus our first function, which is x multiplied by the derivative of the second function, so the derivative of square root of 1 minus x squared with respect to x. And we're going to add the derivative of inverse of sin of x with respect to x. Now let's go ahead and find each derivative. DX divided by DX is basically 1, so we can just cross out that term and we can just say that it is equal to 1. So we get square root of. 1 minus x2 + x multiplied by Notice that square root of 1 minus x 2 is a composite function, so we're going to apply the chain rule. We're going to differentiate. square root of 1 minus x2. With respect to the inner function. 1 minus x 2 and according to the chain rule we're going to multiply by the derivative of 1 minus x2 with respect to x. And finally we're going to add to the derivative of inverse of sin of x. Let's recall that it is equal to 1 divided by. Square root of 1 minus x 2 From here, let's continue. We get square root of. 1 minus x2 + X multiplied by The derivative of the square root of an inner function is going to be equal to 1 divided by 2 square root of the inner function, so we get 1 divided by 2. square root of 1 minus x 2. And we're multiplying by the derivative of 1 minus x 2. With respect to x, the derivative of a constant is 0, the derivative of x 2 is 2x using the power rule. So we're basically multiplying by -2 x. We have that negative sign, right? And we're going to add. 1 divided by the square root of 1 minus x 2. From here we can simplify. 2 divided by 2 gives us 1, right? And continuing the equation, we get square root of. 1 minus x 2 We then have that negative sign. So we're going to write minus x multiplied by x gives us X2. And we are dividing. By Square root Of 1 minus x 2. Followed by Plus One Divided by Square root of. 1 minus x 2 So from here, let's go ahead and notice that we have a common denominator square root of 1 minus x2, meaning we can rewrite our Two fractions or basically a sum of the two fractions as 1 minus x 2 divided by a square root of 1 minus x 2, which is basically Square root of 1 minus x 2, right? We can notice that it gives us plus 1 minus x2 divided by a square root of 1 minus x2, so we have a term divided by the square root of that same term which gives us square root of that term. So we're adding the square root of 1 minus x2. And therefore the derivative is two square roots of. 1 minus x 2 And that's our final answer for this problem. Thank you for watching.

In Exercises 21–48, find the derivative of y with respect to the appropriate variable.37. y=s√(1-s²) + arccos(s)

Watch next

In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
37. y=s√(1-s²) + arccos(s)