Help simplfying an equation

Nov 2012
5
0
United States
Hello everybody,

I'm new here and need a little help. I'm in the process of teaching myself calculus and ran into this function:

y=(a2+x2)(a2-x2)1/2

I was able to differentiate it down to this:

(a2-x2)1/2(a2+x2)+(x(2x(a2-x2)1/2-(x(a2+x2)/(a2-x2)1/2))

The answer to the problem is:

dy/dx=a4+a2x2-4x4/(a2-x2)1/2 My problem is that I don't understand how to get from where I'm at to the correct answer.

Can anyone show me step by step how to get the answer.

Thank you.
 
Dec 2011
2,314
916
St. Augustine, FL.
We are given to differentiate:

\(\displaystyle y=(a^2+x^2)(a^2-x^2)^{\frac{1}{2}}\)

Assuming \(\displaystyle a\) is a constant, we may use the product, power and chain rules as follows:

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

Now, by factoring, we may write:

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

\(\displaystyle \frac{dy}{dx}=\frac{x(a^2-3x^2)}{(a^2-x^2)^{\frac{1}{2}}}\)
 
Nov 2012
5
0
United States
My apologies. When I posted the original function, it should have read:

y=x(a2+x2)(a2-x2)1/2

When I returned to look at what I posted, I noticed it was incorrect.

Thank you, MarkFL2, for your response.
 
Dec 2011
2,314
916
St. Augustine, FL.
Ah, I thought there must be something missing! :) Let's try this again...

We are given to differentiate:

\(\displaystyle y=x(a^2+x^2)(a^2-x^2)^{\frac{1}{2}}=(a^2x+x^3)(a^2-x^2)^{\frac{1}{2}}\)

Assuming \(\displaystyle a\) is a constant, we may use the product, power and chain rules as follows:

\(\displaystyle \frac{dy}{dx}=(a^2x+x^3)\left(\frac{1}{2}(a^2-x^2)^{-\frac{1}{2}}(-2x) \right)+(a^2+3x^2)(a^2-x^2)^{\frac{1}{2}}\)

Now, by factoring, we may write:

\(\displaystyle \frac{dy}{dx}=\frac{1}{(a^2-x^2)^{\frac{1}{2}}}\left((a^2x+x^3)(-x)+(a^2+3x^2)(a^2-x^2) \right)\)

\(\displaystyle \frac{dy}{dx}=\frac{-a^2x^2-x^4+a^4-a^2x^2+3a^2x^2-3x^4}{(a^2-x^2)^{\frac{1}{2}}}\)

\(\displaystyle \frac{dy}{dx}=\frac{a^4+a^2x^2-4x^4}{(a^2-x^2)^{\frac{1}{2}}}\)
 
  • Like
Reactions: 1 person
Nov 2012
5
0
United States
Thank you very much for your help. I truly appreciate it.
 
Dec 2011
2,314
916
St. Augustine, FL.
Glad to assist! :)