Problem

Find the derivative of the function. $ g(x) = (2…

02:21

Problem 42 Hard Difficulty

Find the derivative of the function.
$ y = \sqrt {x + \sqrt {x + \sqrt {x}}} $

Answer

$$
\frac{1}{2 \sqrt{x}}+1\\
\frac{2 \sqrt{x+\sqrt{x}}+1}{2 \sqrt{x+\sqrt{x+\sqrt{x}}}}
$$

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Video Transcript

all right here we have quite the extensive function that we're going to differentiate using the chain rule. And every time I have a square root function, I like to change it to a 1/2 power function before I differentiate. That allows me to use the product or the power rule. Excuse me. So that means I'm going to rewrite this as the quantity X plus the quantity X plus X to the 1/2 to the 1/2 to the 1/2 which really doesn't make anything look better. Um, and it's going to look pretty bad once I take the derivative, but then we'll kind of clean it up after that. So using the chain rule, we have white prime equals. We're going to work on the derivative of the outside function. So the outer 1/2 power function bring down the 1/2 and then take the entire inside and raise that to the negative 1/2. Now we move on to multiply by the derivative of the inside, and this is the insight. So the inside is a some and to find the derivative of a some you find the derivative of each term in the sun. So we start with the derivative of X, and that's one. And now we move on to adding the derivative of the second part of the some, and we're going to need the chain rule for that. So again, a 1/2 power. So bring down the 1/2 and raised the inside to the negative 1/2 and then we need to multiply by its inside derivative. So inside his X plus extra, the 1/2 which is also a some so it's derivative will be the the some of the derivatives. So the derivative of X is one plus the derivative of X to the 1/2 is 1/2 X to the negative 1/2. Now we made it to the very inside of everything. So the next thing we want to do is simplify our answer and change things back into square root signs. So we have this too, which is going to go in the denominator of the entire thing. And because this is to the negative power, it's also going to go in the denominator of the whole thing, and we're going to change it back to a square root sign. So all the 1/2 powers will be changed back to square root signs. So the denominator of the whole big answer is going to have a two in it. And then it's going to have the square root of X plus the quantity X whoops. How about if I go back and change that to a square root as well? I'm gonna make that notation look a little bit better while I'm erasing. Okay, so we're rewriting this thing right here. We're rewriting that. So it's a square root of X plus, the square root of X plus the square root of X. Okay, now we're gonna work on simplifying this thing, and it's our numerator. So it's actually, um Well, we've got the plus one. We can start with that, and then we're gonna work on this part in a very similar way. We just worked on the previous part, so we have the two on the bottom. We have a negative 1/2 power, so that part's gonna go on the bottom as well. So we're going to have the two times the square root of X plus square root X. So that takes care of all of this and then this part will be in the numerator. So in the numerator we have one plus. But now deja vu. This part is going to have a two on the bottom. And because of the negative power, the square roots going to be on the bottom. So one over two square root X So how is that for a tower of incredible numbers and algebra?

Heather Z.

Oregon State University

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 4

The Chain Rule

Problem

Find the derivative of the function. $ g(x) = (2…

Problem 42 Hard Difficulty

Find the derivative of the function.
$ y = \sqrt {x + \sqrt {x + \sqrt {x}}} $

Answer

$$
\frac{1}{2 \sqrt{x}}+1\\
\frac{2 \sqrt{x+\sqrt{x}}+1}{2 \sqrt{x+\sqrt{x+\sqrt{x}}}}
$$

More Answers

Related Courses

Calculus: Early Transcendentals

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Video Transcript

Calculus: Early Transcendentals

Grace H.

Kayleah T.

Caleb E.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

$$\frac{1}{2 \sqrt{x}}+1\\\frac{2 \sqrt{x+\sqrt{x}}+1}{2 \sqrt{x+\sqrt{x+\sqrt{x}}}}$$

Calculus: Early Transcendentals

Grace H.

Kayleah T.

Caleb E.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

Grace H.

Kayleah T.

Caleb E.

Michael J.

$$
\frac{1}{2 \sqrt{x}}+1\\
\frac{2 \sqrt{x+\sqrt{x}}+1}{2 \sqrt{x+\sqrt{x+\sqrt{x}}}}
$$