# Math Help - Did I calculate the 1st and 2nd derivative of this equation correctly??

1. ## Did I calculate the 1st and 2nd derivative of this equation correctly??

question5 on Flickr - Photo Sharing!

The equation is at the link above.
For the 1st derivative I got f' = (49-x^2)^1/2 + (-x)(49-x^2)^-1/2
For the 2nd derivative I got f'' = -98x+x^3 / (49-x^2)^3/2

Are these right?
Thank you

2. The first derivative is right but I think your second one is wrong.

3. Originally Posted by bananasinpajamas
question5 on Flickr - Photo Sharing!

The equation is at the link above.
For the 1st derivative I got f' = (49-x^2)^1/2 + (-x)(49-x^2)^-1/2
For the 2nd derivative I got f'' = -98x+x^3 / (49-x^2)^3/2

Are these right?
Thank you

$f(x) = x \times \sqrt{{49-x^{2}}}$

$f'(x) = [1 \sqrt{49-x^{2}}] + [x \times \frac{1}{2} \times {(49-x^{2})^\frac{-3}{2}} \times (-2x)]
= [\sqrt{49-x^{2}}] - [{x^2} \times {(49-x^{2})^\frac{-3}{2}}]$

$= \sqrt{49-x^{2}} - \frac{x^2}{\sqrt{49-x^{2}}}$

Now use the product rule to find the second derivative. I am a little too lazy topost the complete solution to find the second derivative. Try doing it yourself and post if you have problems. For your convenience, the second derivative looks like this:

$2x \times [\frac{{-x}^2}{2(49-x^2)^\frac{3}{2}} -\frac{1}{2 \sqrt{49-x^{2}}}] - \frac{2x}{\sqrt{49-x^{2}}}$

4. Hello, and thank you for your responses!

I think I got the right f"

f"=2x^3-147x / (49-x^2)^3/2

I also think I got the inflection points x=0, x=-7, x=7

but now I am having a problem interpreting this into the right answer.
I know it can't be D. I'm pretty sure its not A or B because (0,7) is not on this function. So it is B???? Thank you!!!

5. Originally Posted by bananasinpajamas
I'm pretty sure its not A or B because (0,7) is not on this function.
Not the point (0,7) but the open interval (0,7) in x...

6. Originally Posted by tom@ballooncalculus
Not the point (0,7) but the open interval (0,7) in x...

Oh I understand now! Thank you very much to everyone who helped. I really appreciate it.