differentiate-the-function-fxx3-4-x6-2

Question

Differentiate the function.
$f(x)=x^{3}-4 x+6$

Amrita Bhasin · Accepted Answer

this question asked us to differentiate the function F of X is X cubed minus four acts plus sex. Okay, Off Prada vax we take the derivative by moving the exponents to the front and making it the new co Fisher. We then decrease the experiment by one. The derivative of negative for X is the coefficient, which is simply negative, for the derivative of sex is simply zero. Therefore, the answer to this is three X squared minus.

Amy Jiang · Answer

Okay, so we&#x27;re giving the volunteer the falling function, and we actually find its derivative. So that&#x27;s, um, for over six extra power. Three months. Three. Well, that&#x27;s equal to two over. Three extra for three ministry. And now we once f time evaluated at negative two. So that&#x27;s equal to falling to zero. Power of three. That&#x27;s eight. So that&#x27;s two times eight, which is 16/3, minus three. And what is that? Give me putting that into my calculator. I yet 7/3. So we get that f time of negative two. Is he go to 7/3.

Clarissa Noh · Answer

Hi, This is square. So when number six of Stewart&#x27;s file calculus book Onda question asked us to differentiate dysfunction. So first we&#x27;re gonna differentiate both sides d over dear backs three of her four extra, and then we&#x27;re gonna pull out the constant It&#x27;s gonna be 34 Todo dia backs X to that eve and then two to the power rolled. It simplifies to six next to the seventh.

Heather Zimmers · Answer

let&#x27;s find the derivative of this function using the chain rule. So we&#x27;re going to start with the derivative of the outside function. So we have f prime of X equals. The outside function is the fourth power function. So we bring down the four and we raised the inside to the third. Okay, Now we multiply that by the derivative of the inside function, and the inside function is five x to the sixth plus two x cubed, and it&#x27;s derivative would be 30 x to the fifth plus six X squared using thebe power rule. Okay, so the calculus part of this problem is done, and the rest is a bunch of algebraic simplifying and depending on your instructor, they may or may not want you to do a whole lot of simplifying, so I&#x27;ll show you the amount of simplifying I would do. And then I&#x27;ll show you additional steps you could do so that in the event that you see the answer looking different from yours, you can still figure out whether you have something equivalent. So I think what I would do is I would just rearrange the terms and I would write this last one after the after the 1st 1 so I would have four times the quantity 30 x to the fifth power plus six x squared, multiplied by the quantity five x to the sixth power plus two x cubed, uh, cubed. Now that&#x27;s probably not the answer that you&#x27;re going to see in the back of the book. Well, what I might do next is noticed that both of these terms have a common factor of six x squared. So if you factor out that six x squared and you multiplied by the four, you would have 24 x squared times five x cubed plus one. So then you write the rest of it, and that&#x27;s your answer. Now that might also not be the answer that you find in the back of your book or that you&#x27;re given. So what else could we dio? Well, we could notice that this term here inside the cubing function, there&#x27;s a greatest common factor of X cubed. So suppose that we factor out x cubed inside the cubing function were factoring out x cubed. That leaves a factor five x cubed plus two both of those air cubed. Okay, so let&#x27;s see what&#x27;s next. We&#x27;re going to need to get a little bit more space. Okay, so here&#x27;s what we just had. And so we have X cubed, cubed, and we have five x cubed plus two cubed. So let&#x27;s bring that X cubed cubed, which would be X to the ninth out and multiply it by the 24 x squared. So we have 24 x to the 11th power, and then we have our five x cubed plus one. And then we have our five x cubed plus two quantity cubed. So there&#x27;s another form of the answer. And while in my opinion it might not be necessary to take it all the way to that point, it&#x27;s important to know how to in case you need to.

Ernest Castorena · Answer

they give us the function after Lex is equal to X over E to the six X Well, let&#x27;s write this a little. It&#x27;s actually just x e to the negatives execs. So let&#x27;s do product are derivative is equal Teoh. But I take the derivative of the first factor, not just the one. And then we leave the second factor alone. Then do the opposite this time Leave the first factor alone. Take the derivative of the second factor. So we&#x27;ll have a negative 600. And if we factor out, are you to the negative six x We see that our function just either. The negatives XX times one minus six I was

Ernest Castorena · Answer

this time they give us the function is equal to you. Six over this great of X squared plus four and then we&#x27;ll take the derivative to get. Let&#x27;s see. This is input to the negative one of power so we can pull out the negative 1/2 to the front. Well under with the negative three and now to become the input to the detriment. Negative 1/2. Well, it&#x27;s negative 3/2 times the derivative of the input. But that&#x27;s just to acts on the enemy and up with Let&#x27;s see, negative six x over the square root of X squared plus four. But all of that should be cute.

Problem

Differentiate the function. $f(t)=\frac{1}{2} t^…

Problem 7 Easy Difficulty

Differentiate the function.
$f(x)=x^{3}-4 x+6$

Answer

$f^{\prime}(x)=3 x^{2}-4$

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James Stewart

Biocalculus Calculus for the Life Sciences

Catherine R.

Caleb E.

Kristen K.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

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$f^{\prime}(x)=3 x^{2}-4$

Biocalculus Calculus for the Life Sciences

Catherine R.

Caleb E.

Kristen K.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

James StewartBiocalculus Calculus for the Life Sciences

Catherine R.

Caleb E.

Kristen K.

Joseph L.

James Stewart

Biocalculus Calculus for the Life Sciences