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Find the derivative of the function using the definition of a derivative. State the domain of the function and the domain of its derivative.$G(t)=\frac{4 t}{t+1}$

domain of the function $\mathrm{R}-[1]$domain of the derivative $\mathrm{R}-[1]$

Calculus 1 / AB

Chapter 3

Derivatives

Section 2

The Derivative as a Function

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Okay. So again, we want to use the function, the definition of the derivative of a function to solve the derivative and so that given a function, F uh, the derivative is defined as the limit as each approaches zero of the function evaluated at X plus H minus, the function evaluated at each over H I went ahead and made a typo says, minus the functioning violated at X. Okay. And so for this instance, we are given our function. F of tea is going to be equal to 40 over t plus one, and this means that are functioning. Evaluated at T plus H would be equal to 40 plus four age over t plus H plus one so we can plug these into our definition, which gives us the limit as H approaches zero of 40 plus four h over T plus age plus one minus four t over T plus one all that divided by H. Okay, so what we want to do is multiply the fracture on the left by t plus one over t plus one and the one on the right by Teeples H plus one over T plus age plus one. Okay, So when we do that, we will end up with the limit as each goes to zero of 40 square plus 40 plus for H T plus four H minus for a T squared minus for th minus 40 over t plus H plus one times t plus one times h. So now we'll get some cancellations, right? So our four teas will cancel the four times T times h will cancel, and the four t squared will cancel. So all that's left is on top is for H, and so this age will cancel with E H on the bottom. Okay, so it gives us the limit as age goes to zero of four over T plus age plus one times t plus one. So if we evaluate this at H equals zero, we get that this is equal to four over t plus one squared. Okay, so that is the solution for the derivative. And I want to take a look at the domain. So if we started the domain of the function here, the only problem is that we don't want the denominator to be equal to zero. So in this case, our domain is going to be negative. Infinity to you. Negative one with negative one to infinity. Right? Or you could consider it all real numbers minus so all real numbers, but without negative one. And in fact, we get the exact same domain for the derivative, right? Anything except for negative one will evaluate here. So for both the function and the derivative, our domain is all real numbers, with the exception of the number negative one.

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