# Help simplfying an equation

#### Crisor2431

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.

#### MarkFL

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:

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

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

#### Crisor2431

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.

#### MarkFL

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:

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

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

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

• 1 person

#### Crisor2431

Thank you very much for your help. I truly appreciate it.

#### MarkFL

Glad to assist! 