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An alternative way to do implicit differentiation is to introduce a mythical extra variable $t .$ Think of $x$ and $y$ as both functions of $t .$ Now, use the fact that $\frac{d y}{d t}=\frac{d y}{d x} \frac{d x}{d t}$(a) Solve for $d y / d x$. (b) For the equation $x^{2}+y^{3}=6,$ find the relationship between the rates of change of $x$ and $y$ as in the previous section. (c) Use the result of parts (a) and (b) to find $y^{\prime}$.

(a) $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$(b) $-\frac{2 x}{3 y^{2}}$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 10

Related Rates

Derivatives

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Boston College

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.

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$ x$ and $y$ are functions…

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Use implicit differentiati…

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$x$ and $y$ are functions …

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Find $y^{\prime}=\frac{d y…

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this problem we're going toe Look at implicit differentiation in a slightly different way. We're going to start with the fact that I can say that d y d t equals d Y d x times dx DT Okay, we're going to say that X and Y are both functions off t at So first thing we want to dio given this piece of information, let's solve for D y DX that gives us d y d t divided by dx DT. And if you think about these were just numbers I can multiply top and bottom by D. T. And that would make sense that this algebraic lee This looks correct. But we're going to use that piece of information on a problem. And the equation we're gonna look at here is why squared plus our sorry x squared plus y cubed equals six. Now we're told that X and Y are both functions of tea. So just like we've been doing with these related rate problems, we're going to take the derivative of both of these with respect to t. So that would become two x dx DT plus three y squared d Y d t and the derivative of a constant is just zero. Hey, and I would probably put one of these on one side and one on the other, but for right now, that's sufficient. We've taken the derivative of both of these with respect to t. Now, it kind of seems like we've done two separate problems here, but we're gonna put them together. I'm gonna combine what we did here with what we just found by taking the derivative with respect to T. And I want to find d y d X. And we're gonna do that by fighting this quote shit. So let's let's put one on one side. Three y squared d y d t equals negative two x the x DT So just moving that piece over now I want to find d y d t over dx DT. So first, I'm going to divide both sides by three y squared. Okay, that's going to give me d y d t equals negative two x over three y squared. And now I will divide both sides by d x d t. So that gives me negative two x over three y squared. But this piece that left hand side do I Dhe said it. I'd write it wrong. That is the right hand side of what? We found it first so that whole fraction is really D y d x. So we've used this imaginary t to take the derivative and solve for D Y D X. If you notice our final answer doesn't have t anywhere in it because our original equation didn't really have tea. We just put that in as a concept, something to help us find our answer. So we can kind of combine our implicit differentiation and what we've been doing with related rates to find another way to take the derivative.

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