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Finding derivative of a square root - Quick question!

  • Thread starter nukeman
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  • #1
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Homework Statement



Ok, working on a inverse function question, and I got stuck with something.

Can someone explain the steps that makes this possible here. Something I am missing :(

f(x) = √(x^3 + x^2 + x + 1)

How is the inverse of the above function this...

3x^2 + 2x + 1 / 2√(x^3 + x^2 + x + 1)


Homework Equations





The Attempt at a Solution



I just looked at f(x) = √(x^3 + x^2 + x + 1) and went:

1/2(3x^2 + 2x + 1)^-1/2

Thats as far as I got...:(
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement



Ok, working on a inverse function question, and I got stuck with something.

Can someone explain the steps that makes this possible here. Something I am missing :(

f(x) = √(x^3 + x^2 + x + 1)

How is the inverse of the above function this...

3x^2 + 2x + 1 / 2√(x^3 + x^2 + x + 1)


Homework Equations





The Attempt at a Solution



I just looked at f(x) = √(x^3 + x^2 + x + 1) and went:

1/2(3x^2 + 2x + 1)^-1/2

Thats as far as I got...:(
You aren't finding the inverse, you are taking the derivative. The part you are missing is the chain rule. (u^(1/2))'=(1/2)u^(-1/2)u'. Put u=x^3 + x^2 + x + 1. It's a special case of (f(g(x))'=f'(g(x))*g'(x).
 
  • #3
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You aren't finding the inverse, you are taking the derivative. The part you are missing is the chain rule. (u^(1/2))'=(1/2)u^(-1/2)u'. Put u=x^3 + x^2 + x + 1.
SORRY! Mistake... yes I am not finding the inverse. See below:

f(x) = √(x^3 + x^2 + x + 1)

How is the DERIVATIVE of the above function this...

3x^2 + 2x + 1 / 2√(x^3 + x^2 + x + 1)
 
  • #4
Ray Vickson
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Homework Helper
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Homework Statement



Ok, working on a inverse function question, and I got stuck with something.

Can someone explain the steps that makes this possible here. Something I am missing :(

f(x) = √(x^3 + x^2 + x + 1)

How is the inverse of the above function this...

3x^2 + 2x + 1 / 2√(x^3 + x^2 + x + 1)


Homework Equations





The Attempt at a Solution



I just looked at f(x) = √(x^3 + x^2 + x + 1) and went:

1/2(3x^2 + 2x + 1)^-1/2

Thats as far as I got...:(
What you wrote is not the inverse, but the derivative (but badly written). What you wrote reads as
[tex] 3 x^2 + 2x + \frac{1}{2 \sqrt{x^3 + x^2 + x + 1}},[/tex]
but perhaps you wanted to write
[tex] \frac{3x^2 + 2x + 1}{2 \sqrt{x^3 + x^2 + x + 1}}.[/tex] In that case you should use parentheses, like this:
(3x^2 + 2x + 1)/(2 √(x^3 + x^2 + x + 1)).

The inverse of f(x) is much harder to get, as it would involve the solution of a 4th degree polynomial.

RGV
 
  • #5
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0
What you wrote is not the inverse, but the derivative (but badly written). What you wrote reads as
[tex] 3 x^2 + 2x + \frac{1}{2 \sqrt{x^3 + x^2 + x + 1}},[/tex]
but perhaps you wanted to write
[tex] \frac{3x^2 + 2x + 1}{2 \sqrt{x^3 + x^2 + x + 1}}.[/tex] In that case you should use parentheses, like this:
(3x^2 + 2x + 1)/(2 √(x^3 + x^2 + x + 1)).

The inverse of f(x) is much harder to get, as it would involve the solution of a 4th degree polynomial.

RGV
You are correct, sorry.
 
  • #6
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,543
755

Homework Statement



Ok, working on a inverse function question, and I got stuck with something.

Can someone explain the steps that makes this possible here. Something I am missing :(

f(x) = √(x^3 + x^2 + x + 1)

How is the inverse of the above function this...

3x^2 + 2x + 1 / 2√(x^3 + x^2 + x + 1)
That isn't the inverse function. What it looks like is ##\frac{dy}{dx}## or ##f'(x)##.

Homework Equations





The Attempt at a Solution



I just looked at f(x) = √(x^3 + x^2 + x + 1) and went:

1/2(3x^2 + 2x + 1)^-1/2

Thats as far as I got...:(
If you are calculating ##f'(x)## you aren't done. You need the chain rule (the derivative of the "inside"). Noting that you haven't actually stated the problem you are working on, let me guess that it has something to do with the derivative of the inverse function, not the inverse function itself. What is the relationship between the derivative of a function and the derivative of its inverse?

[Edit] Wow! All those replies while I was typing. I must really be slow...
 
  • #7
Dick
Science Advisor
Homework Helper
26,258
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SORRY! Mistake... yes I am not finding the inverse. See below:

f(x) = √(x^3 + x^2 + x + 1)

How is the DERIVATIVE of the above function this...

3x^2 + 2x + 1 / 2√(x^3 + x^2 + x + 1)
I told you how. The numerator is u', the 2 in the denominator comes from (1/2) and √(x^3 + x^2 + x + 1) is u^(-1/2). Match them up with the parts of (u^(1/2))'=(1/2)u^(-1/2)u'. Chain rule.
 
  • #8
655
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Thanks Dick, just got it!

I got confused as to why the 3x^2 + 2x + 1 went on top.
 
  • #9
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With over 600 posts here at PF you shouldn't be writing this:
3x^2 + 2x + 1 / 2√(x^3 + x^2 + x + 1)
when you mean this:
$$ \frac{3x^2 + 2x + 1}{2\sqrt{x^3 + x^2 + x + 1}}$$

If using LaTeX is too much for you, at least use enough parentheses so that what you write can be clearly understood, like this:

(3x^2 + 2x + 1) / (2√(x^3 + x^2 + x + 1))
 

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