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Question

Find the derivatives of the given functions.
$$v=3\left(t+\ln t^{2}\right)^{2}$$

Tyler Moulton · Accepted Answer

we want to find dy dx for the function Y is equal to three times X plus L n x squared square This question is testing your knowledge of differentiation, shortcuts or transcendental functions like logarithmic functions as such. We need to note relevant rules to solve. On the left of Rule zero derivatives for logarithmic functions, lobby you and Ellen you on the right. I have the power of product rule and channel effectively in this problem we need to combine rule zero and rule three to solve. So why is already written in its most easily differential form? We can apply the chain rule to the function in parentheses that is squared and then particularly what&#x27;s inside to solve. So do I. D. S is by the chain rule three times two times X S L n x square to the first power times of what&#x27;s inside which is one plus one over X squared times two X. One over x squared two X two over X. So this simplifies to the following. do I d x equals six times X plus l n x squared times one plus two over X as X box in the bottom right of the solution?

Amy Jiang · Answer

Okay, so we want to take the derivative of a fallen function. So here, and use the course a little here. But let&#x27;s not do that instead, that would simplify. Will we have correctly have so splitting this up into three factions? We have one bus on three over T shirts to her negative one and plus you to the car later, too. Now let&#x27;s dig prohibitive. He&#x27;s doing it this way. It&#x27;s much simpler. We get negative three. She took our negative, too. One in minus two. T two. Power of negative three. Okay, so we get the following for our interpretive.

Heather Zimmers · Answer

in this problem, we&#x27;re finding the derivative of two to the power t cubed. So let me remind you that if you have an exponential function to to the X, it&#x27;s derivative will be two to the X Times natural log tube. So if we apply the chain ruled to that and we have an exponential function to to the f of X. It&#x27;s derivative will be two to the f of X times natural log two times the derivative of the inside, which would be f prime of X. So that&#x27;s the situation we have going on here. So the derivative will be to to the t cubed times natural log to times the derivative of the inside function. So times the derivative of Tea Cube, which is three t squared

Bobby Barnes · Answer

we want to find the derivative of the team is able to t squared minus one over T squared plus team minus two. So you could just start by using the caution war to take the derivative of this. But before we do that, let&#x27;s see if we can back this to possibly supply a little bit. So our numerator will too t plus one t minus one difference of squares and our denominator here. Well, we ask yourselves what number can multiply a negative to or what? Seven numbers won&#x27;t find negative too. But after one Principe&#x27;s t cost you over T minus one man, you might notice that he&#x27;s t minus ones or counts. And now we have t plus one over t plus two, which significantly simplifies what we&#x27;re trying to do here. And this one thing you need to make sure to write up onto the side that t cannot equal to what So I&#x27;ll take the drivel about t plus one of the teapots too. So, like I was saying, this is crucial. And so just to remind you of the suits so quotient rule. So the way I remember is the following. So ddx of some high function over some No function. So it&#x27;s slow. D Hi minus. Hi. Hello. All over the square of what&#x27;s below. So in this case, our highest people&#x27;s wanna Arlo is t plus two so we can Hey, t x t x there&#x27;s gonna bea t in this case dd t oh t cost one over t plus two. Well, this is going to be low. So maybe it&#x27;Ll be a good idea if I read these in different colors. Oh, all right. I like the new male Blue will be team plus one on the denominator green tea plus two so we can take low first. So t cross too. Team of what we have in the numerator t plus one minus what we have in the new mayor t plus one time to drive. They were respected T t too all over what we have in the denominator squared two plus two square. So the derivative of t plus one and keep us too Well, both of these are in the form out a derivative of t course from constant C, so we could go ahead and use some outstanding property to first break us. You team plus T c. The derivative of a constant is zero and a derivative of T to the first power will just be one. Since we could move the one out front using power rule and their team, you should try that. Get tea to zero and then teach zero is just what? So each of these derivatives here and become one So have I t plus two times one minus t plus one times one over t lost too square No. Now T minus t is going to counsel out and then two minus one is one So we can rewrite this as one over t plus two squared And then just recalling that this function should not be defied at T minus one and likewise are derivative. Should also be not to find that tea is equal to what So this is our derivative and t does not equal one is just one of the other values. This derivative is not supposed

Alexander Wittmond · Answer

Oh, okay. So to divert derivative here, we need to use every precaution, rule or the product rule. I prefer to use a private room. So the first thing we could do is pull out of the two, and then we have tea to the three house prime over Ellen of two t to the three house plus one plus t to the three halves. Times one over. Ellen times two to tee to the three halves, plus one prime. Now differentiating we have two are here, then three halves and hear T to the one half times Ellen of two t to the three halves plus one minus sense. Over here, we&#x27;re differentiating one over X on the outside. So you get t to the free house over Ellen of to teach of a free house, plus one squared times the derivative of the inside, which, in his case, is Ellen to t to the three house plus one. Private. Okay, Now allow me to Oops. Not what I wanted to do, but close. Clear up some clutter. Uh, let&#x27;s move this a little bit forever. Not quite so much. Okay, Now give me back my turn. Okay? Now we have two times three teach of Ah, one half services. This is one half. So that&#x27;s the same thing as that&#x27;s the same thing as just rooty here. Then we have to Ellen of to tee to the three house three house, um, close one minus t to the free house. Time&#x27;s over. Ellen of two t to the three house plus one squared times one over, um, two times t to the free house, three halves, plus one times the derivative of the insight here, which is just going to be free root tea. Okay, finally, we just need to simplify, and we can multiply in this, too. To get three root tea times Ellen of two t to the free has plus one times to t to the three house plus one minus six route T all over. Two times I learned of two t to the free house plus one, um, all squared times to tee to the three halves, plus one

Amy Jiang · Answer

Okay, so you see that we have a product of two terms. So let&#x27;s use our product role in taking our derivative that&#x27;s using or taking the driver of a three team Nice one that&#x27;s equal to three times have fallen. And then plus the derivative of this term here. So we use our power. Well, let&#x27;s think one to T minus two. Power of native to sometimes derivative over inside. And that&#x27;s to So he gets three over to Chief minus two and then minus two over to T minus two. Power to. Okay, so we get the following, as are derivative.

Problem

Find the derivatives of the given functions. $$y=…

Problem 15 Easy Difficulty

Find the derivatives of the given functions.
$$v=3\left(t+\ln t^{2}\right)^{2}$$

Answer

$\frac{d v}{d t}=\frac{6(t+2) \ln \left(t+\ln t^{2}\right)}{t}$

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$\frac{d v}{d t}=\frac{6(t+2) \ln \left(t+\ln t^{2}\right)}{t}$

Basic Technical Mathematics with Calculus

Catherine R.

Heather Z.

Kristen K.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Catherine R.

Heather Z.

Kristen K.

Michael J.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus