Example Problem 4. Implicit Differentiation

Implicit Differentiation. Example Problem 1 Find this, d squared y over dx squared when x equals negative 1 and y equals negative 2.So Leibniz's notation for the secondderivative is, in essence, finding f'(x) even though it is not a function, so we cannot call it f'(x). In order to find the second derivative of the function, we must first find the first derivative. The derivative of x squared is2x, and the derivative of y squared is 2y, but we must take intoaccount the differential dy dx in order to find that the derivative of 5 is 0. Now, at this point, we do not want to take the second derivativeagain. We have not officially found the first derivative. We onlyhave an equation that involves the first derivative; so we need to isolate or solve for dy dx; so subtract 2x and then divide the 2y,you end up getting negative 2x over 2y; and do a littleforethought here and reduce 2 divided by 2 to be 1. So, dy dx isnegative x over y. Now we can take the derivative again. Take the derivative with respect to x, take the derivative with respect tox of that, so the derivative of the derivative, that's the secondderivative. And then the derivative of this is a quotient. And so we go low times the derivative of high, that's negative 1 minus high negative x times the derivative of low, derivative ofy, that's dy dx, all over low squared. x was negative 1, y was negative 2.So we can start evaluating dsquared y, dx squared is equal to y. If it's negative, 2 times negative 1. We can find the derivative of x²y² by returning to our definitionof the derivative and noticing that dy/dx is equal to negativex/y. Since we said that x equals negative 1, y equals negative 2, and dx equals 1, then dy/dx is equal to -1/2. So multiply by negative 1/2, all divided by 2 squared - negative 2squared, and then we get the distinct pleasure of having to cleanall this up. So 2 plus 1/2 over 4, so that gives us 2 and 1/2, so 5 over 2is equal, over 4, which would be 5 over 8.There's the second derivative.