# Thread: Differentiation involving Confusing Cube Roots

1. ## Differentiation involving Confusing Cube Roots

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.

2. 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.

Hi AeroScizor,

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

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

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

3. 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.

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

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

5. 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?

6. 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?
$\displaystyle \displaystyle\left(\frac{a\sqrt{x}}{a^2+x^2}\right )^{-\frac{1}{2}}=\frac{1}{\sqrt{\frac{a\sqrt{x}}{a^2+x ^2}\right)}}}=\sqrt{\frac{a^2+x^2}{a\sqrt{x}}}$

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

7. Originally Posted by Sudharaka
$\displaystyle \displaystyle\left(\frac{a\sqrt{x}}{a^2+x^2}\right )^{-\frac{1}{2}}=\frac{1}{\sqrt{\frac{a\sqrt{x}}{a^2+x ^2}\right)}}}=\sqrt{\frac{a^2+x^2}{a\sqrt{x}}}$

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

8. (Happy)Hello, AeroScizor!

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

I rewrote the function:

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

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

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

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

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

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

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