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A demand equation is given by $p=\sqrt{9-x}$, where $p$ is the price per item when $x$ items are demanded. Find $d R / d x$ when $x=1$.

$$\frac{15 \sqrt{2}}{8}$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 8

Implicit Differentiation

Derivatives

Campbell University

Harvey Mudd College

University of Nottingham

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.

03:06

For the demand equations, …

00:39

At what value of $x$ does …

01:24

Where $x$ represents the q…

05:06

Demand function, $p=D(x),$…

02:47

02:15

If a price-demand equation…

01:08

Assume that the demand fun…

01:55

02:12

for this problem. We've been given a function. P equals the square root of nine minus X, and we're told that P is the price per item, and X is the number of items that are being demanded. Now, what we're looking for here is actually a derivative not of P or of X, but of our I want to find d r d X at the point where X equals one. So how do I find the revenue function? The revenue function is x times p the number of units times the price per unit. No, in this case, I have X and then P is the square root of nine minus x. So in order to find this derivative, I actually don't need to use implicit differentiation at all. I could just use the product rule that says it's the first times the derivative of the second. So I have nine minus X square root is to the half power. So bring down that half. That puts a two in my denominator subtracting one from the exponents. Give me negative one half, which puts that radical in the denominator. And then I multiplied by the derivative of what's under my radical sign, which is a negative one. So first times, the derivative of the second plus the second times, the derivative of the first, which is just one. So there's my derivative. It's not terribly nice looking, but that's okay, because what I don't really what I really want to do with this is evaluate it. So I'm going to come up here and I'm going to take my derivative and evaluated were X equals one. So if I let X equal one, that first term is gonna be negative 1/2. And that means I would have the square root of eight. And it's the same radical, so it's gonna be plus the square root of eight. So now let's do some algebra. Eight is, um, could be simplified because it is has a factor of four, so I can next equals one. I can change that square root of eight into two times the square root of two that becomes four in the denominator, and this is to square root of two. Now, I'd like to rationalize my denominator, so I'm going to multiply top and bottom of this first term here by square root of two that gets rid of that radical in my denominator. So I have negative square root of two over eight and then plus two squared of to. Well, I'd like to put these together, So let's do a common denominator. Multiply top and bottom by eight, and that gives me 15 square root of two over eight. So that is the derivative evaluated. The point X equals one.

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