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13. Determine the equation of the tangent line to In(xy) + y = Zat the point (e,1)....

Question

13. Determine the equation of the tangent line to In(xy) + y = Zat the point (e,1).

13. Determine the equation of the tangent line to In(xy) + y = Zat the point (e,1).



Answers

Find the equation of the line tangent to the function at the given point. $$f(x)=1 / x^{2} \text { at }(1,1)$$

For the given problem we want to find the equation of the line tangent to the function at the given point. So we're gonna have f of X. Yeah equals one over X squid. Um Well we know that one over X squared is X to the negative too. So when we take the derivative we get a -2 X to the -3. If we plug in X equals one, We're not just getting a -2 as our slope. So our line is going to be y equals -2. X must be, We know the .1. 1. Okay, is in question. So when X equals one, let's copy this over here. So we're gonna have x equal to one And why equal to 1? So negative two plus B equals one. We solve for B by adding two on both sides. So B equals three. Since the equals three. We can write this as y. It was a negative two X plus three and that's our final answer.

Okay, so a function here is Sfx is equal to one over X plus one. So to find the derivative here, S prime of X. Using the definition, well we want to basically the definition of a function F X. If we have F prime of X is gonna equal to the limit. As a joke goes to zero of F of X plus age minus the functions or minus F of X. All divided by H. So let's go to first here let's find s of X plus H. So S of X plus age. That's going to be equal to one over well X plus H plus one. So he's putting in an X plus H for X and our function. And then um the derivative here is going to be equal to the limit as a church goes to zero of well, S of X plus H minus s of X. That's going to be one over X plus H plus one minus one over X plus one. So minus the function minus one over X plus one. And then all divided by H. So now we just find a common denominator here of X plus one times X plus H plus one. Um And then we get the limit as a church tends to zero of what we end up with here with a X plus one minus X plus H plus one. Um Divided by X plus one times that common denominator of X plus one times X plus H plus one. And then this is that all divided by H. So we have here the limit As a joke goes to zero all that seat. Um While we get an X plus one minus X minus h minus one. Everything cancels except for a minus H. So we have a minus H over the X Plus one times x Plus H Plus one. Um And then we're still this is that again still divided by H. Um But this is the same thing. The same negative H. Over X. Plus while negative H over X plus one times X plus eight plus one times one over H. Is divided by H. So this H really can be put here in the denominator and then we have while negative H over H cancels to just leave a negative one. Therefore we have the limit as H Goes to zero up negative one over um X plus one times X plus H. Plus one. Okay? But as H goes to zero, well this age goes to zero. We just have negative one over X plus one times X plus one. Which is equal to just um negative um negative one over X plus one times X plus one is X plus one quantity squared. Therefore the derivative here s prime of X is equal to negative one. Over the quantity X plus one squared

So whenever we want to find the equation of a tangent line, it's important to think about how do you write the equation of a line. What's most helpful is the point slope form of a line, which is why minus Y. One equals M X minus X. One. And we already know a given why value is 1/5. And for that why we know X is one. So really all we need to do is find the slope and that instantaneous slope is called a derivative. So how are we going to find the derivative of our function? Well, since it's a quotient, something divided by something else, let's use the quotient rule. So the derivative which I'll denote as Y prime is the derivative of the numerator times the denominator unchanged and minus the derivative. The denominator with the numerator unchanged. And then all divided by that denominator but squared, Well the derivative of one is 0. So that eradicates that first term and the derivative of X squared plus four would be two X plus zero. So we get negative two x times one over X squared plus four. But squared When we plug in our known x value of one We get negative two times 1 In one squared plus four is 5 but squared is 25 and that's our slip. So going back to our point slope form, we know the equation of our line are tangent line is why -1 5th M Now is negative 2 25th An X -1. So with the quotient rule in the point slope form of the line were able to find the equation of the tangible. Um

In the problem we have three X Squire. Why plus X. Q. White ass plus Y. Que plus X. into three. Why Squire? Why does that equal zero now XQ plus three X. Or the Squire? Where does become -3? Expo square wine minus? Like you r minus equal to minus three x. esquire y less. Like you bond X cubed plus three X. Y. Square. So we have, why does add one Common 3 equal to -9 Upon 7? Hence the equation. Uh huh. Detainment line become why am minus? Why not? So why not? is three That equals m. Which is -9.7 X -X Not is one. So this is the occasion of dependent line, hence it is dancer.


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