Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$A(x)=2\sqrt{x^2-16}+\frac14\sqrt{68x^2-x^4-256}\;,\;\; (4 < x < 8)$$

of which the derivative is:

$$a'(x)=\frac{2x}{\sqrt{x^2-16}}+\frac{136x-4x^3}{8\sqrt{68x^2-x^4-256}}$$

I first had to find a value of $x$ for which $A'(x)=0$ The result I got was a local maximum at $x=7.1296$ and a local minimum at $x=4$ The part I am struggling with is:

Verify by the second derivative test that this value of $x$ corresponds to a local maximum of $A(x)$ .

As I understand I have to find the derivative of the above function, then find the derivative of that and then input $x=7.1296$ to solve the equation and if the answer is $<0$ then the local maximum is at $x=7.1296$ . Please let me know if I have the correct method. If so, then I'm struggling to get the second derivative and need help with that.

share|improve this question
    
The lack of necessary parenthesis makes this incomprehensible. Are you taking the square root of the whole of $x^2 - 16$, $68x^2 - x^4 - 256$ or something else? –  Lee Yiyuan Feb 25 at 14:23
    
You can use dollar symbols ("$") to input mathematical formulas. See this: math.stackexchange.com/editing-help#latex –  frabala Feb 25 at 14:31
    
My apolgies I was having trouble entering it. I have edited the function and it is now correct. –  skeeto Feb 25 at 15:29
    
I am not sure whether this helps, but maybe some parts of this expression could be simplified a little using $x^4-65x^2+256=(x^2-4)(x^2-64)$. –  Martin Sleziak Feb 25 at 15:34

1 Answer 1

Yes, your sketched approach is just fine. I'll write your first derivative using the exponent $1/2$ to replace the squareroot symbol, and I also factored out $4$ from the numerator to simplify.

$$A'(x) = \dfrac{2x}{(x^2-16)^{1/2}} + \frac {34x - x^3}{2(68x^2 - x^4 - 256)^{1/2}}$$

You might want to try combining the two terms by finding a common denominator. Then to find $A''(x)$, you'll only need to use the quotient rule once.

$$A'(x)= \frac{4x(68x^2-x^4-256)^{1/2}+(34x-x^3)(x^2 - 16)^{1/2}}{2\Big((x^2-16)(68x^2 - x^4 -256)\Big)^{1/2}}$$

It's hard to say whether using the quotient rule twice will be easier than using the quotient rule once but needing the product rule a number of times. Either way, it may seem very messy, but you need to simply persevere, being careful along the way. (Personally, I'd go with the "original" $A'(x)$.

Hang in there, and feel free to check back if you need any verification.

share|improve this answer
    
The function you've taken is incorrect which was probably an error on my part in the way I typed it. I have edited the function and it appears as it should now –  skeeto Feb 25 at 15:30
    
Note: I simply factored out $4$ from the numerator of the second term in your sum: $$136x-4x^3 = 4(34 x - x^3)$$ $$A'(x)=\frac{2x}{\sqrt{x^2-16}}+\frac{136x-4x^3}{8\sqrt{68x^2-x^4-256}} = \frac{2x}{\sqrt{x^2 - 16}} + \frac{\color{blue}{4}(34x - x^3)}{\color{blue}{8}\sqrt{68x^2 - x^4 - 256}}$$Then, $\frac 48$ reduces to $\frac 12$. –  amWhy Feb 25 at 15:34
    
Also note that $\sqrt{f(x)} = (f(x))^{1/2}$. So "my" A'(x) is precisely the same A'(x) you posted (before, and now). I made out your earlier notation just fine. ;-) –  amWhy Feb 25 at 15:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.