- #1

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Differentiate:

g(x)=√(4-x^4) , x is a set of [-√2, √2] and determine the domains.

So I got the derivative which is,

g'(x)=(-4x^3) ^1/2

What should I do with the -+√2 ?

I don't know what to do next

- Thread starter CrossFit415
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- #1

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Differentiate:

g(x)=√(4-x^4) , x is a set of [-√2, √2] and determine the domains.

So I got the derivative which is,

g'(x)=(-4x^3) ^1/2

What should I do with the -+√2 ?

I don't know what to do next

- #2

Mark44

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Nope. You need to use the chain rule. It will help to write your function asI'm on mobile so I can't use latex.

Differentiate:

g(x)=√(4-x^4) , x is a set of [-√2, √2] and determine the domains.

So I got the derivative which is,

g'(x)=(-4x^3) ^1/2

g(x) = (4 - x

What should I do with the -+√2 ?

I don't know what to do next

- #3

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Alright

- #4

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I have

g'(x) = 1/2(4-x)^-1/2 • -4x^3

Then how would I determine the domains? I know 4-x^4 >= 0. So what about the two square roots?

What should I do with the +/-√2 ?

g'(x) = 1/2(4-x)^-1/2 • -4x^3

Then how would I determine the domains? I know 4-x^4 >= 0. So what about the two square roots?

What should I do with the +/-√2 ?

Last edited:

- #5

Mark44

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Since your original is named g, its derivative is g', not G'.

The domain of g is the interval [-[itex]\sqrt{2}[/itex], [itex]\sqrt{2}[/itex]]. The domain of the this derivative will be exactly the same, with the possible exception of the endpoints.

- #6

HallsofIvy

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NO. Do it again and show each step.I have

G'(x) = 1/2(4-x)^-1/2 • -4x^3

The domain will be the intersection of the natural domain of the derivative (not what you have above) and the given domain of the function, [itex][-\sqrt{2}, \sqrt{2}][/itex].Then how would I determine the domains? I know 4-x^4 >= 0. So what about the two square roots?

What should I do with the +/-√2 ?

- #7

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g'(x) = 1/2(4-x^4)^1/2-1 • d/dx (4-x^4)

= 1/2(4-x^4)^-1/2 • -4x^3

Should I keep continuing?

Then

= -2x^3(4-x^4)^-1/2

= 1/2(4-x^4)^-1/2 • -4x^3

Should I keep continuing?

Then

= -2x^3(4-x^4)^-1/2

Last edited:

- #8

HallsofIvy

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Please do NOT use "x" both as the variable and for multiplication!

- #9

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Sorry!Please do NOT use "x" both as the variable and for multiplication!

- #10

Mark44

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Much better.g'(x) = 1/2(4-x^4)^1/2-1 • d/dx (4-x^4)

= 1/2(4-x^4)^-1/2 • -4x^3

Should I keep continuing?

Then

= -2x^3(4-x^4)^-1/2

When you write expressions on a single line, you need to use more parentheses. The above should be written as -2x^3(4-x^4)^(-1/2).

- #11

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Thanks! Ohh ok I ll be sure to use more parenthesis next time.Much better.

When you write expressions on a single line, you need to use more parentheses. The above should be written as -2x^3(4-x^4)^(-1/2).

So should I just plug in +/- √2 for x to find the domains? Or leave it alone? Are the domains just +/- √2 ?

- #12

Mark44

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Your question was answered twice in this thread.So should I just plug in +/- √2 for x to find the domains? Or leave it alone? Are the domains just +/- √2 ?

I don't think you understand what "domain" means - it's not domains. The domain (singular) is the set of numbers at which the relevant function is defined.

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