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

• Mar 27th 2010, 12:02 PM
bananasinpajamas
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(Happy)
• Mar 27th 2010, 12:25 PM
dwsmith
The first derivative is right but I think your second one is wrong.
• Mar 27th 2010, 12:37 PM
harish21
Quote:

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(Happy)

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

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

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

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:

$2x \times [\frac{{-x}^2}{2(49-x^2)^\frac{3}{2}} -\frac{1}{2 \sqrt{49-x^{2}}}] - \frac{2x}{\sqrt{49-x^{2}}}$
• Mar 28th 2010, 11:36 AM
bananasinpajamas
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!!!
• Mar 28th 2010, 12:06 PM
tom@ballooncalculus
Quote:

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...
• Mar 28th 2010, 12:38 PM
bananasinpajamas
Quote:

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. (Clapping)