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Let f(x)=x ($\sqrt{x}+\sqrt{x+9}$). Question is to check if it is differentiable or not at x=0. Edit : FOR ABOVE FUNCTION I HAVE TO CHOOSE OUT FOLLOWING 1. continously differentiable at x=0

  1. continous but not differentiable at x=0

  2. differentiable but derivative is not continous at x=0

  3. not differentiable at x=0

My attempt I differentiated, so f '(x)=x($\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+9}}$) +$\sqrt{x}+\sqrt{x+9}$. I see IT IS DIFFERENTIABLE at x=0 (I applied definition to be sure) option1 is correct . i hope i am not wrong

Is above process correct?

BigBang
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  • Expand and simplify the beginning of what you wrote for $f'(x)$ – Claude Leibovici Feb 06 '15 at 11:34
  • What you derive is $f'(x)$ for $x>0$. If you want to determine the right derivative of $f$ at $0$, you can either determine $\lim_{x\to0+}f(x)/x$ directly, or appeal to the theorem that that $\lim_{x\to0+}f'(x)$ exists and $f$ is right continuous at $x=0$ implies that $f_+'(0)$ exists. – Yai0Phah Feb 06 '15 at 12:11
  • @FrankScience does existence of this one sided limit sufficient to imply that f is differentiable at x=0 – BigBang Feb 06 '15 at 12:30
  • @BigBang It should be one-sided differentiable. I've mentioned, though. – Yai0Phah Feb 06 '15 at 13:02

2 Answers2

1

You have derived $$f '(x)=x(\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+9}}) +\sqrt{x}+\sqrt{x+9}$$ which you could simplify to$$f '(x)=\frac32\frac{x+6+\sqrt{x}\sqrt{x+9}}{\sqrt{x+9}}.$$

Now note that $f'(x)$ evaluated at zero is $3$.

Math-fun
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  • your algebra is wrong: it should be $\sqrt{x(x+9)}$ in the denominator. – Alex Feb 06 '15 at 12:19
  • I checked again, it does not look like that I made a mistake ... – Math-fun Feb 06 '15 at 12:25
  • $\frac{\sqrt{x+h} -\sqrt{x}}{h} \to_h \frac{1}{2 \sqrt{x}}$ it's not differentiable at $x=0$. – Alex Feb 06 '15 at 13:53
  • on $x>0$ is differentiable, no? well at zero it is not differentiable since the left derivative does not exist as mentioned earlier. thanks for the comment. But why is the simplification I wrote for $f'(x)$ wrong? (thanks for the comment) – Math-fun Feb 06 '15 at 14:43
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It's enough to notice that $\sqrt{x}$ is not defined for $x<0$, therefore it can't be continuous at $x=0$, therefore it can't be differentiable at this point.

Alex
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  • is it necessarily discontinous bcoz of that? shouldn't we discuss only right hand limit at x=0? – BigBang Feb 06 '15 at 12:05
  • Continuous means two-sided limit exists and the function at the point is equal to its limit. If either of the conditions fails, the function is not continuous. – Alex Feb 06 '15 at 12:10
  • Look at this definition: http://en.wikipedia.org/wiki/Continuous_function#Definition_in_terms_of_limits_of_functions – Alex Feb 06 '15 at 12:12
  • i was in doubt, bcoz i thought that at boundry points we discuss only one sided limit. – BigBang Feb 06 '15 at 12:14
  • right-hand limit exists, that's true – Alex Feb 06 '15 at 12:14
  • http://math.stackexchange.com/questions/560307/prove-that-sqrtx-is-continuous-on-its-domain-0-infty – Alex Feb 06 '15 at 12:15
  • I'd say (4) is the best option, because 1) and 2) don't mention 1-sided continuity – Alex Feb 06 '15 at 12:18
  • atleast we can say it is continous. sure? – BigBang Feb 06 '15 at 12:21
  • No: it's right-continuous. 'Continuous' is a much stricter condition, it requires existence of the limit on both sides, look at the wikipedia link I gave you and the similar question. – Alex Feb 06 '15 at 12:22
  • both sided limit is for interior points. here x=0 is not interior point of domain of this function. so only one sided continuity is valid, i think. but what about differentiablility? – BigBang Feb 06 '15 at 12:25
  • Are you familiar with the definition of differentiability? – Alex Feb 06 '15 at 13:53