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(a) Find $ y' $ by implicit differentiation. (b)…

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Problem 3 Medium Difficulty

(a) Find $ y' $ by implicit differentiation.
(b) Solve the equation explicitly for y and differentiate to get $ y' $ in terms of $ x. $
(c) Check that your solutions to part (a) and (b) are consistent by substituting the expression for $ y $ into your solution for part (a).

$ \sqrt{x} + \sqrt{y} = 1 $

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Video Transcript

In this problem were given an equation, wording equation is of x, plus the root of y is equal to 1. In part, a we asked to use implicit differentiation to find very to function y. So let's stood out. Let'S take derivative of all the terms with respect to x. Preston would give us 1 over 2 square root of x. Plus second term would give us 1 over 2 square. Now we're in to inner function since y is also a function of have dx and that is equal to 0, because 1 is just a constant. So let's leave the i d x on the 1 side we have y t is then equal to negative 1. Over 2 star root of x, divided by 1 over 2 square root. I means to would cancel out and you would find the answer as the negative of y divided by s x now in part b, were asked to explicitly solve for y and then take derivative of that function. So we see that square root of y is equal to 1 minus square root of x. Now, if we take square of sides, we find y to be 1 minus square root of x. Squared now, let's find which of this y prime is equal to 2 times 1 minus square root of x times the roof inner function, which is negative 1 over 2 square root of x and 2, will cancel out and we find then the answer to be negative. So square root of x, minus 1 give i by square root of x all right now in part c, we are asked to compare the derivative we find in part and part b. So, let's take derivative of the function y that we find in part a so we have negative of the root of y divided by the square root of x. Now, in part, we find what our function y is, and from this we see that y is equal to negative of square root of 1 minus square root of x square square, root, divided by square x, square root and square will cancel out and will end up With negative of 1 minus square root of x, divided by square root of x, and that is equal to square root of x, minus 1 divided by square and as you can see, this is same as what we had for. So we can see that you can say that both implicit and explicit differentiation gi us the same result.

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Calculus: Early Transcendentals

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