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$$5 s^{2}\left(v^{3}-1\right)=7 . \text { Find }(a) d s / d v;(b) d v / d s$$.

(a) $\frac{2\left(1-v^{3}\right)}{3 s v^{2}}$(b) $\frac{3 s v^{2}}{2\left(1-v^{3}\right)}$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 8

Implicit Differentiation

Derivatives

Campbell University

Harvey Mudd College

Baylor University

Idaho State University

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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for this problem. We want to take our given equation and find the derivative, and we're going to do it in two ways. We're going to find D S D V and DVDs. In other words, D STV means that s is a function of V DVDs. Means that V is a function of s. So let's do these one at a time D S d v So s is a function of V, which means every time I see hates the derivative of S, I'm going to need to attack on a D S d. V because of the chain rule. So I'll take the derivative of s is I normally would. But since s is a function of V. I have to put that d S d V in there as well. So let's look at our equation. Five s squared times V cubed minus one equal seven. Take the derivative of both sides will have a product. So we need the product rule. That is the first times the derivative of the second minus the second times the derivative of the first, which is 10 s times D S. D V. There's that Shane. While we were talking about. And the derivative of a constant is just zero. Okay, now let's do some algebra. Anything that is a Oops. I'm sorry. That's a plus in the middle. Not a minus. My apologies. Product rules. A plus. Anything is a D S. D. V. Term is going to stay on the left hand side. Anything that's not is gonna go on the right. So what I have is 10 s times V cubed, minus one D S T V. As for the equal negative five s squared times three v squared. Now, I could do a little bit of canceling here. I have one. The left side as an s, the other side as an s square. So I could get rid of one of the essence. And each one has a factor of five. So I could get rid of the five on the right and the 10 becomes a two on the left Final step I'm going to solve for D S D. V. So that gives me negative s. Actually, let me get rid of these parentheses. Negative three s v squared, and my denominator. I'm gonna have a to and then V cubed minus one. So that is the derivative D S. D. V. Now, what about the other way DVDs? Well, in that case, V is a function of s. So any time I take the derivative of V, the changeable says I need to put in a D v D s. It's the same concept, just off of a different letter. Okay, so once again, I'm going to copy our equation. Once again, we take the derivative, we have a product. So that is first times the derivative of the second. Okay. Again, um, that's in this case. I do need to put in the D V. D s. Because V is a function of s the way we're looking at it right now. So first times, the derivative of the second plus the second times, the derivative of the first, which is 10 s Now, this time we don't need to put on anything else, because s is the variable. And this is going to equal zero. So just like before, let's see what we can cancel. I haven't s on both sides. One side has a five or one piece has. So I'm jumping the gun. Let me get rid of that. I know going to canceling, but let's actually solve it first. So let's put a pieces on both sides. On the right hand side, I have five s squared times three V squared DVDs on the now on the right hand side, I'll have negative 10 s times V cubed, minus one. Now let's you're canceling. Sorry I jumped the gun. They're a bit get rid of one s on each side. The five goes in there once goes in there twice. And when I solve, I get D V D s on the top. I'll have negative, too. The cubed minus one. And the denominator will be three s the squared. So those there are two derivatives D S, D. V and D V D s.

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