In Exercises 65 and 66, find the derivative using the definiti...

Question

In Exercises 65 and 66, find the derivative using the definition.

ƒ(t) = 1 .

2t + 1

Accepted Answer

Welcome back, everyone. In this problem, using the definition of the derivative, we want to calculate the derivative of the function G of Y equals 1 divided by 4Y + 5. A says it's 1 divided by 4 Y + 5 squared, B-1 divided by 4Y + 5 squared. C 4 divided by 4Y + 5 squared, and the D-4 divided by 4Y + 5 squared. Now how are we going to use the definition of the derivative to calculate the derivative of the function? Well, this is referring to the limit definition of the derivative, and basically what it tells us is that the derivative of g of Y. Is going to be equal to the limit as it approaches 0 of G of Y plus H minus G of Y. All divided by H. So essentially, if we can find G Y plus H, because we already have GFY, then we'll be able to plug it into this expression or into this limit definition to help us find the derivative of GFY. Now, here, to find G Y plus H, all we need to do is to plug Y + H wherever we see Y in the function G. So G Y + H is going to be 1 divided by 4 multiplied by Y + H+ 5. Now that we have g of Y plus H, then we can find our derivative. Cuz no, this means our derivative is going to be equal to the limit. As H approaches 0 of 1 divided by 4 multiplied by Y plus H + 5. Minus 1 divided by 4 Y + 5, OK, all divided by H. Now to simplify this expression, let's try to combine both of the fractions in our numerator. When we do that, We're going to get the limit as H approach is 0 of 4 Y + 5, OK, minus 1 multiplied by 4 multiplied by Y + H+ 5, OK. I know this is all going to be divided by 4 multiplied by Y plus H + 5, OK. Multiplied by 4 Y + 5, OK. Multiplied by H because remember our entire expression was being divided by H. Now here, notice we can simplify what's in our numerator, OK. So, no. Not here, if we expand or -1 here, it's gonna be 4 Y plus 5, minus 4 multiplied by Y4Y, minus 4 multiplied by H4H minus 5. All divided by our expression in the denominator. Now do you see any way to make our numerator simpler to simplify? But here, 4 Y minus 4Y equals 05 minus 5 equals 0, so we're left with negative 4H and our numerator. So this is the limit as it approaches 0 of -4H. Divided by 4 multiplied by Y + H+ 5. Multiplied by 4 Y + 5. Multiplied by H. Now we can cancel it from the numerator and denominator, and now we can directly substitute 0 for wherever we see it. And now when we do that, we're gonna have -4 in our numerator and in our denominator, if H is 0, then this is gonna be Y + 0 or just 4Y. So in our denominator, we're gonna have 4 Y plus 5 multiplied by 4 Y + 5. And then we can rewrite that as -4 divided by 4 Y plus 5 squared. Thus, the derivative of G of Y is equal to -4 divided by 4 Y + 5 squad, which means D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

In Exercises 65 and 66, find the derivative using the definition.

ƒ(t) = 1 .
2t + 1

Key Concepts

Derivative Definition

Limit

Function Composition

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