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Problem 34 Hard Difficulty

(a) If $ f(x) = x + 1/x $, find $ f'(x) $.
(b) Check to see that your answer to part (a) is reasonable by comparing the graphs of $ f $ and $ f' $.

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Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 2

Limits and Derivatives

Section 8

The Derivative as a Function

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Video Transcript

this problem. Number thirty four of the story. Calculus. In addition, section two point eight. Party if ever Vex equals X plus one Rex, find F Primex. So let's use the definition one before the derivative. This is the limit. And experts, eh? Of the function ethnics minus every guided by X minus eighty. Okay, our limit will be FX X plus one of rex minus the function evaluated, eh? A plus one over, eh? All divided by X minus a. Okay, Right Next step will be to multiply by and the lowest common denominator to eliminate the fractions. So an axe over eggs limited express his aim. Um X times x x squared plus one over x Times X A minus eight times x His critics minus one toe Ray Tim's a X minus X all over X minus. Hey, time's a X Now, awaken groups some of the terms in the numerator. That's group the X squared and the cortex term. Let's factor out that X and we're just up with explaining saying on their many terms are minus the quantity expensing and all this divided by, uh, explain. It's a in the pine by X Now, each turn in. The Raider has an expense. A turn I can cancel with e expectancy in the denominator. And that leaves us with a X minus one surrounded by a ex as expertise. Ain't this becomes he squared minus one threated by a squared or one minus one over a squared. And if we generalize is for any a value in the domain of the function Given its that I mean being any number X, then our dirt is one minus one over x squared. So a scored executed first to the same derivative function. So we've concluded the dirt over this function is one minus one over X Quit now Bernie's ah, the grafts of infinite Pregnant To confirm that the derivative of dysfunction experts when Rex is indeed one man is one of the X word. So here the two functions planted on the same graph the blue is the original function affects and oranges that order function. If we look at the blue function, we see that there are straight lines here to this straight line or we'LL have a constant slope. Therefore a concentrated because it slips that the tension lines or constant. So that's what we see here on the dirt of graph. It's a constant value exactly equal to one. So it's constant until it reaches a point where this slope against change a little bit and then we have, ah, local movement, Maxima, which has the slip of zero. That's what we see here constant slope and then it decreases to zero. And then afterward we have a very sharp decrease in the slope of the tension line, so very negative slips. And this derivative function shows very negative soaps as we approach Sarah from the left. As we approach zero from the right for this function. After Becks, we see that the slopes of the tension lines are very, very negative. So this continues Armand to the right of zero very, very negative slopes until we reach this local minimum, which has a slip of zero that is consistent with this derivative or function. And then afterwards the slopes increase until the slopes become a concept once again. So disturbing a function is definitely consistent with our original function and confirms that we computed that our function correctly

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The unemployment rate $ U(t) $ varies with time. …

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Calculus: Early Transcendentals

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Catherine Ross

Heather Zimmers

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Catherine Ross

Heather Zimmers

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Grace He

Catherine Ross

Heather Zimmers

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals