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$1 / t+1 / s=1 .$ Find $(a) d s / d r ;$ (b) $d t / d s,$ (c) Show that $(d s / d t)(d t / d s)=1$.

(a) $-s^{2} / t^{2}$ or $\frac{s-1}{1-t}$(b) $-t^{2} / s^{2}$ or $\frac{1-t}{s-1}$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 8

Implicit Differentiation

Derivatives

Campbell University

Oregon State University

Baylor University

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.

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for this problem. We've been given a function one over T plus one over s equals one, and we're going to find two different derivatives. We're gonna find the derivative of s with respect to t and the derivative of tea with respect to s. So in other words, for one, we're gonna let it be a function of t the other one. We're going to be a function of s. So let's start with our first case de s DT. Now I'm going to rewrite our equation here instead of having denominators. Let's write this with exponents. This is T to the negative. First plus s to the negative first equals one. I just find that a little bit easier to take our derivatives when their exponents rather than quotations. Okay, so let's start with D S. D T. That means that, uh, s t is my variable and s is a function of t. So when I take this derivative tease my variable So I could just take this as a normal derivative negative t subtract one to the negative, too. Now, how about as to the negative one? Well, that's going to be negative s to the negative too. But chain rule says I have to now take the derivative of s with respect to t. And that's going to equal zero. Let's move That t squared over. So I have negative s to the negative two d s d t equally it positive t to the negative too. Yeah, and d s d t t to the negative to over negative s to the negative too. I don't really like toe leave the negative exponents there, so I'm going to change each of them to a positive by flipping its position. That becomes negative to the positive Two on top, over T to the positive two on the bottom. So that is D S d t derivative of s with respect to t. Now let's do it the other way d t d s. So I'm going to start with the same same function. I just took this one right over here. Now, when I take the derivative t is, um yeah, sorry s is the variable. I'm taking the derivative of tea with respect to s. So now when I take this derivative negative t to the negative too G t d s minus s to the negative, too. But that one I just leave equals zero and again, just like last time. I'm going to solve for the derivative. So I'm gonna bring over my term without the d t. D s. That becomes as to the negative, too. And I'm going to divide by negative t to the negative too. And again, I like positive exponents. So I'm going to flip their spots and have negative t squared over s squared and something you can see from these if you look these air reciprocal of each other. So if I take d s d t times d t over d s Look what we get, We get negative s squared over t squared times negative t squared over s squared. Well, everything cancels and it equals one

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