University:
Rice University
Course:
MATH 212 | Multivariable Calculus
Academic year:

2023
Views:

71

Pages:

2
Author:

ChromeApril

Implicit Differentiation. Example Problem 2 Let's look at another implicit derivative question. The derivative of the function can be found by taking the derivative of each of its components. We first take the derivative of x cubed and plug in 2 for x and 1 for y: . The derivative of y cubed is 3y squared, so when we plug in our values for x and y. Derivative of negative x is negative 1. To isolate dydx, we simply need to add the two-step rule for derivatives of exponential functions and apply it to the given expression. The derivative of 10 with respect to x is equal to 10x0. The goal is still to solve for dy dx, or using the quadratic formula, 3 times 2 squared minus 3 times 2 times negative 1 squared equals 0. So 2 squared is 4 times 3 is 12. The negative one-squared is positive, because it has a plus sign after it. Minus 6 times the integral of negative one-third x squared is going to be positive 13, because the negative one-third times another negative number gives you a positive number. Multiplying by a negative one-third gives you a positive number. Divide by negative six and you get an answer of 13 over 6.