Differentiation involving Confusing Cube Roots

**AeroScizor** · March 5th 2011, 08:29 PM

Hi all,
Attached in the post are the questions:

Been struggling with the first one for quite a while because there is one cube root inside another. So hopefully someone can help me out with that.

The second question is bugging me as I am unsure whether to use the quotient rule(seeing it is in the form if a fraction) or the chain rule. I would appreciate if someone could offer some advice on what to do and maybe tips on which rule to use in what scenario.

Thanks in advance. Your help will be greatly appreciated.

**Sudharaka** · March 5th 2011, 11:05 PM

Originally Posted by AeroScizor

Hi all,
Attached in the post are the questions:

Been struggling with the first one for quite a while because there is one cube root inside another. So hopefully someone can help me out with that.

The second question is bugging me as I am unsure whether to use the quotient rule(seeing it is in the form if a fraction) or the chain rule. I would appreciate if someone could offer some advice on what to do and maybe tips on which rule to use in what scenario.

Thanks in advance. Your help will be greatly appreciated.

Hi AeroScizor,

For both of these problems you have to use the chain rule and the quotient rule. In the first problem, take $u=\displaystyle\frac{a\sqrt{x}}{a^2+x^2}$

$\displaystyle\frac{d}{dx}\sqrt{\frac{a\sqrt{x}}{a^ 2+x^2}}=\frac{d\sqrt{u}}{du}\times\frac{du}{dx}=\f rac{1}{2\sqrt{u}}\frac{du}{dx}$

To find $\dfrac{du}{dx}$ you will need the quotient rule.

**Pulock2009** · March 5th 2011, 11:33 PM

Originally Posted by AeroScizor

Hi all,
Attached in the post are the questions:

Been struggling with the first one for quite a while because there is one cube root inside another. So hopefully someone can help me out with that.

The second question is bugging me as I am unsure whether to use the quotient rule(seeing it is in the form if a fraction) or the chain rule. I would appreciate if someone could offer some advice on what to do and maybe tips on which rule to use in what scenario.

Thanks in advance. Your help will be greatly appreciated.

see attachment for answer. differentiate the ln terms, the constant =0 and the a2+x2 term that should be easy

**HallsofIvy** · March 6th 2011, 03:28 AM

Are these supposed to be cube roots or square roots? You say "cube roots" but your picture shows square roots.

**AeroScizor** · March 11th 2011, 03:57 AM

Originally Posted by HallsofIvy

Are these supposed to be cube roots or square roots? You say "cube roots" but your picture shows square roots.

Apologies it was a typo error. I am ashamed to say that I still don't quite grasp the concept. I understand how to apply the chain rule to bring down power 1/2 in the first question and also how to apply quotient rule in the later stage. However, from there on, I am stumped as to how to further simplify the function, specifically how do I simplify the (a square root x/ a^2+x^2)^-1/2 term. Would appreciate if someone could guide me along. Thanks

P.S: How do you all type mathematical functions and values?

**Sudharaka** · March 11th 2011, 06:14 AM

Originally Posted by AeroScizor

Apologies it was a typo error. I am ashamed to say that I still don't quite grasp the concept. I understand how to apply the chain rule to bring down power 1/2 in the first question and also how to apply quotient rule in the later stage. However, from there on, I am stumped as to how to further simplify the function, specifically how do I simplify the (a square root x/ a^2+x^2)^-1/2 term. Would appreciate if someone could guide me along. Thanks

P.S: How do you all type mathematical functions and values?

$\displaystyle\left(\frac{a\sqrt{x}}{a^2+x^2}\right )^{-\frac{1}{2}}=\frac{1}{\sqrt{\frac{a\sqrt{x}}{a^2+x ^2}\right)}}}=\sqrt{\frac{a^2+x^2}{a\sqrt{x}}}$

The mathematical text are typed by LaTeX commands. You can read some posts in the Latex help section to get some idea. http://www.mathhelpforum.com/math-he...orial-266.html

**AeroScizor** · March 12th 2011, 05:01 PM

Originally Posted by Sudharaka

$\displaystyle\left(\frac{a\sqrt{x}}{a^2+x^2}\right )^{-\frac{1}{2}}=\frac{1}{\sqrt{\frac{a\sqrt{x}}{a^2+x ^2}\right)}}}=\sqrt{\frac{a^2+x^2}{a\sqrt{x}}}$

The mathematical text are typed by LaTeX commands. You can read some posts in the Latex help section to get some idea. http://www.mathhelpforum.com/math-he...orial-266.html

Cheers that was a lifesaver

**Soroban** · March 12th 2011, 06:46 PM

(Happy)Hello, AeroScizor!

$\text{Differentiate: }\;f(x) \;=\;\sqrt{\dfrac{a\sqrt{x}}{x^2+a^2}}$

I rewrote the function:

. . $\displaystyle f(x) \;=\;\left(\frac{ax^{\frac{1}{2}}}{x^2+a^2}\right) ^{\frac{1}{2}} \;=\;\frac{a^{\frac{1}{2}}x^{\frac{1}{4}}}{(x^2+a^ 2)^{\frac{1}{2}}} \;=\;\sqrt{a}\cdot\frac{x^{\frac{1}{4}}}{(x^2+a^2) ^{\frac{1}{2}}}$

$\displaystyle\text{Then: }\;f'(x) \;=\;\sqrt{a}\cdot\frac{(x^2+a^2)^{\frac{1}{2}}\cd ot\frac{1}{4}x^{\text{-}\frac{3}{4}} - x^{\frac{1}{4}}\cdot\frac{1}{2}(x^2+a^2)^{\text{-}\frac{1}{2}}\cdot2x}{x^2+a^2}$

. . . . . . . . . . $\displaystyle=\;\sqrt{a}\cdot\frac{\frac{1}{4}x^{\ text{-}\frac{3}{4}}(x^2+a^2)^{\frac{1}{2}} - x^{\frac{5}{4}}(x^2+a^2)^{\text{-}\frac{1}{2}}}{x^2+a^2}$

Multiply numerator and denominator by $4x^{\frac{3}{4}}(x^2+a^2)^{\frac{1}{2}}$

. . . . . $\displaystyle f'(x) \;=\;\sqrt{a}\cdot\frac{(x^2+a^2) - 4x^2}{4x^{\frac{3}{4}}(x^2+a^2)^{\frac{3}{2}}} \;=\;\frac{\sqrt{a}\left(a^2 - 3x^2\right)}{4x^{\frac{3}{4}}(x^2+a^2)^{\frac{3}{2 }}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

When differentiating a fraction to a power: . $y \:=\:\left[\dfrac{f(x)}{g(x)}\right]^n$
. . I find it easier to differentiate: . $y \;=\;\dfrac{[f(x)]^n}{[g(x)]^n}$

(I seem to be the only one on this planet that teaches that.)

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