# Thread: differentiate y= [(x^3 + 2)^1/2][(x^3 +7)^-1/2]

1. ## differentiate y= [(x^3 + 2)^1/2][(x^3 +7)^-1/2]

I used the chain rule first for each of the two functions (g(x) and f(x)), getting

g'(x) = 3/2x^2(x^3 + 2)^-1/2

f'(x) = -3/2x^2(x^3 + 7)^-3/2

Then, using the product rule, I get f(x)g'(x) + f'(x)g(x) =

= [(x^3 +7)^-1/2][3/2x^2(x^3 + 2)^-1/2] + [-3/2x^2(x^3 + 7)^-3/2] [(x^3 + 2)^1/2]

which is just a mess, and I don't know how (or if it's possible) to simplify it.
Is that correct so far?
Ta!

2. ## Re: differentiate y= [(x^3 + 2)^1/2][(x^3 +7)^-1/2]

You can keep going,

Organise the first 2 brackets into one fraction and do the same for the last two brackets, take the lowest common denominator which should be:

$\displaystyle 2(x^3+2)^{\frac{1}{2}}(x^3+7)^{\frac{3}{2}}$

3. ## Re: differentiate y= [(x^3 + 2)^1/2][(x^3 +7)^-1/2]

Originally Posted by canger
I used the chain rule first for each of the two functions (g(x) and f(x)), getting

g'(x) = 3/2x^2(x^3 + 2)^-1/2

f'(x) = -3/2x^2(x^3 + 7)^-3/2

Then, using the product rule, I get f(x)g'(x) + f'(x)g(x) =

=[(x^3 +7)^-1/2][3/2x^2(x^3 + 2)^-1/2] + [-3/2x^2(x^3 + 7)^-3/2] [(x^3 + 2)^1/2]

which is just a mess, and I don't know how (or if it's possible) to simplify it.
Is that correct so far?
Ta!
Hi canger,

Yes it is correct.

$\displaystyle \left[(x^3 +7)^{-\frac{1}{2}}\right]\left[\frac{3x^2}{2}(x^3 + 2)^{-\frac{1}{2}}\right]+\left[-\frac{3x^2}{2}(x^3 + 7)^{-\frac{3}{2}}\right] \left[(x^3 + 2)^\frac{1}{2}\right]$

Since $\displaystyle \frac{3x^2}{2}$ and $\displaystyle (x^3 +7)^{-\frac{1}{2}}$ are common factors you can take it out of the brackets and write separately.

$\displaystyle =\frac{3x^2(x^3 +7)^{-\frac{1}{2}}}{2}\left\{(x^3 + 2)^{-\frac{1}{2}}-(x^3 + 7)^{-1}(x^3 + 2)^\frac{1}{2}\right\}$

$\displaystyle =\frac{3x^2(x^3 +7)^{-\frac{1}{2}}}{2}\left\{\frac{1}{(x^3 + 2)^{\frac{1}{2}}}-\frac{(x^3 + 2)^\frac{1}{2}}{(x^3 + 7)}\right\}$

$\displaystyle =\frac{3x^2}{2(x^3 +7)^{\frac{1}{2}}}\left\{\frac{1}{(x^3 + 2)^{\frac{1}{2}}}-\frac{(x^3 + 2)^\frac{1}{2}}{(x^3 + 7)}\right\}$

Now take the lowest common denominator of the two fractions inside the brackets and simplify it. Hope you can do it from here.

4. ## Re: differentiate y= [(x^3 + 2)^1/2][(x^3 +7)^-1/2]

It looks like you just take out the $\displaystyle (x^3 + 7)^-1/2$
Is that right?

Thank you so much!

P.S. How did you get your power of (-1/2) look right?

5. ## Re: differentiate y= [(x^3 + 2)^1/2][(x^3 +7)^-1/2]

Originally Posted by canger
It looks like you just take out the $\displaystyle (x^3 + 7)^-1/2$
Is that right?

Thank you so much!

P.S. How did you get your power of (-1/2) look right?
Yes I didn't notice that. Since $\displaystyle (x^3+7)^{-\frac{1}{2}}$ is also a common factor it could be taken out of the brackets also. I have corrected that in my previous post.

$\displaystyle (x^3+7)^{-\frac{1}{2}}$-------["tex"](x^3+7)^{-\frac{1}{2}}["/tex"] I hope now you understand how to write the fractional power correctly.