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Find $d y / d x$ using any method.$$x^{2 / 3}+y^{2 / 3}=1$$

$$-\frac{y^{1 / 3}}{x^{1 / 3}}$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 8

Implicit Differentiation

Derivatives

Campbell University

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

04:36

Find $d y / d x$ using any…

02:00

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00:42

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01:32

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02:24

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01:55

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00:41

05:01

01:02

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00:58

Find $\frac{d y}{d x}$, i…

for this problem, we have the function X to the two thirds power, plus why to the two thirds power equals one. And we want to find the derivative of our function with respect to X. So we could do this either by solving for why, using an explicit differentiation or we could take the derivative just as it is implicitly with our X's and y's altogether. We're going to do it that way, mostly because that's what this lessons been about. The biggest thing to remember when you're doing implicit differentiation is that why is a function of X So I'm taking the derivative of why, for example, here, in just a moment, it'll be Why did the two thirds, When I take the derivative of that, I'm going to do it just like I would for any variable, I bring down the exponents, subtract one, but why isn't just a variable. Why is a function? So the chain rule says after I do that, uh, exponents rule there, I need to take the derivative off. Why of that inside function. So I'm gonna have to multiply by d Y d X every time we take a derivative with y in it when we're doing implicit differentiation, I end up with a d Y d X. So let's go back to our function here X to the two thirds. Let's just a plain X So that derivative looks exactly like what we've been doing up to this point derivative of why this is where that chain rule comes into play. So I have to take take the derivative and then I multiplied by D Y d X, and the derivative of one is just zero. Well, first of all, I'm gonna multiply every term both sides by three halves that gets rid of both of my, uh, coefficients. And now I'm going to solve for D Y d x The term without d y dx goes over to the right hand side and I'm gonna divide by why to the negative one third. Now, I don't wanna leave all those negative exponents. Remember, negative exponents could be made positive by pushing it either from the numerator to the denominator or vice versa. So the why will become a positive one third by putting it in the numerator theme, X becomes a positive one third by putting it in the denominator. So here is the derivative for my given function

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