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Question

Find the derivative of each function.
$$f(x)=\sqrt{\left(x^{2}+1\right)(\sqrt{x}+1)^{3}}$$

Ernest Castorena · Accepted Answer

this time they want us to take the derivative of the following function we have. Rex is equal to the square of X squared, plus one times the square root of X plus one. Cute. Yeah, And so we end up with Let&#x27;s see, let me take our derivative. We have 1/2 times the input X squared plus one Times square directs plus one cube all of that to the negative 1/2 power and I&#x27;m gonna multiply by literally of the inside. But that&#x27;s gonna be a product will well to x. Then leave the second Tamerlan X squared plus one cubed. And this time, do the reverse. Leave the first term alone, but take the derivative of the second. This one will require General square root of X plus one. That&#x27;s the inputs word and times the derivative of that. Well, what&#x27;s in? Let&#x27;s put that way in the friend. So we&#x27;ll have close one Health X to the negative one. Help! Yeah, this is our derivative. We could simplify it, but it looks good

Ernest Castorena · Answer

Vega was the function Apple X is equal to the square root of X over X squared plus one. Well, lets him take their a tive on the to do 1/2 of the input. So enough times X x crib plus one and then document the power. It&#x27;s not a negative one. But now when you take that area of the inside and this will require a question room the ruler of the top times the bottom so x squared plus one and minus this time leave the top alone times the derivative of the bomb, We&#x27;ll end up with two X. Does that look right? Mm. Yeah, they&#x27;re the bottom is two X and not all over the bottom square. Well, this is minus two X squared, so end up with a one minus x squared on the number, and that isn&#x27;t I hate cancels with anything in the bottom so we could leave our and say like this. Um, it looks good.

Ernest Castorena · Answer

I get to this problem? They give us the function. Let&#x27;s see, Scary of X cubed plus two those two x all of that to the negative to part. So let&#x27;s do Chainarong take their this so input to the negative to power will have negatives you times thean But this time to the negative three power And now when you take the derivative of the inside. But that is just like why now? Times x cubed plus two to the negative 1/2 power. But this was getting a general too because we have three x squared for the input and then don&#x27;t forget the other term we have ah, last two. Yeah. Now this looks good.

Ernest Castorena · Answer

for this problem. They give us the function that Lex is equal to you. X over the square of X squared, plus one. And now we&#x27;re going to need to take the derivative solid scene. We&#x27;ll have the riveter with top multiplied by the bottom. So x squared, plus one to the one hour power. Then we&#x27;ll subtract this time, leaving the top alone but taking the derivative of the bottom so well how one, huh? Let&#x27;s see X squared plus one to the negative one. How and then all of this over, we&#x27;ll square the bottom so we&#x27;ll have X squared plus one. Actually, was there anything missing from that? Let&#x27;s see, We also have to multiply by the input at the top when we take the director that a nominator in the second time. Well, I want to buy my drinks currency well enough where it has an X squared plus one in the denominator. So if we&#x27;re gonna get common denominators, we should multiply the top and bottom by that, maybe also by two. No, no, this 1/2 kinds of that with the TV here. So under with an X squared plus one over the square e of our experience plus one. And this also has that in the denominator. But it also has an X squared up top. So we have a minus exported all over that and then all over an X squared plus one. So what do we enough with we end up with when these kinds of out one over X squared plus one? Well, one and then 1/2. So a three over to part on the model nothing.

Rakesh Kumar Sharma · Answer

it is what? She They&#x27;re ready for the from excess. Where? Kiss my Mr dont buy food dollars So let&#x27;s not simplify. This morning we find this country be being accessed about Cuba. Yes, Jean, By just doing who existing one minus one layer is different. Your genetics was that he would get with the physician off this with me. You know the extension? Yes, your positional discipline. Do you? Do you exercise one minus up? Okay. Deposition on the stormy. Where your stool Even minus half. You exist if I my if you follow this. If I think so, but the position will be todo do exist in my house for this. Well, the second number The transition existing one minus huh Way Continue. Totally getting plus existing Bottom line that is this.

Ernest Castorena · Answer

the following function. And while let&#x27;s just make it easier for us to take the power, we&#x27;ll have three x cubed minus that C x to the 1/2 plus one. The text to the three over to our Now let&#x27;s see, we take power. Alter my term or a nine expert on multiplied by the three. Bring that they have sent to the front. So we&#x27;ll get minus three halfs X to the one, huh? Doesn&#x27;t.

Problem

Find the derivative of each function. $$f(x)=\sqr…

Problem 20 Easy Difficulty

Find the derivative of each function.
$$f(x)=\sqrt{\left(x^{2}+1\right)(\sqrt{x}+1)^{3}}$$

Answer

$\frac{(\sqrt{x}+1)^{2}\left[4 x^{\frac{3}{2}}(\sqrt{x}+1)+3\left(x^{2}+1\right)\right]}{4 \sqrt{x} \sqrt{\left(x^{2}+1\right)(\sqrt{x}+1)}}$

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Robert T. Smith, Roland B. Minton

Calculus: Early Transcendental Functions

Grace H.

Catherine R.

Heather Z.

Caleb E.

Precalculus Review - Intro

Functions on the Real Line - Overview

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$\frac{(\sqrt{x}+1)^{2}\left[4 x^{\frac{3}{2}}(\sqrt{x}+1)+3\left(x^{2}+1\right)\right]}{4 \sqrt{x} \sqrt{\left(x^{2}+1\right)(\sqrt{x}+1)}}$

Calculus: Early Transcendental Functions

Grace H.

Catherine R.

Heather Z.

Caleb E.

Precalculus Review - Intro

Functions on the Real Line - Overview

Robert T. Smith, Roland B. MintonCalculus: Early Transcendental Functions

Grace H.

Catherine R.

Heather Z.

Caleb E.

Robert T. Smith, Roland B. Minton

Calculus: Early Transcendental Functions