Derivative definition in 0

[Nivelo]

Question: By the definition the derivative of f(x)=( (x^2+6)^1/2 )^1/2=sqrt(sqrt(x^2+6)) in 0 is a limit f'(0)=lim h->0 Q of a special quota Q. Decide the quota.

I calculated the derivative and got f'(x)=x/(2*(x^2+6)^(3/4) and then f'(0)=0/2*6^(3/4)=0. So now the equation is

0=lim h->0 Q, this is where i am lost, what is it that im supposed to calculate?

 

[ksdhart2]

The question's worded pretty weirdly and it's making me unsure, but it seems like you did a whole lot of unnecessary work in even calculating the derivative. The exercise is about the definition of the derivative, so it makes sense to start there:

\(\displaystyle \frac{dy}{dx} = f^{\prime}(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\)

You're then asked about the derivative at the point x = 0, so it follows that:

\(\displaystyle f^{\prime}({\color{red}0}) = \lim_{h \to 0} \frac{f({\color{red}0} + h) - f({\color{red}0})}{h} = \text{???}\)

 

[Dr.Peterson]

Question: By the definition the derivative of f(x)=( (x^2+6)^1/2 )^1/2=sqrt(sqrt(x^2+6)) in 0 is a limit f'(0)=lim h->0 Q of a special quota Q. Decide the quota.

I calculated the derivative and got f'(x)=x/(2*(x^2+6)^(3/4) and then f'(0)=0/2*6^(3/4)=0. So now the equation is

0=lim h->0 Q, this is where i am lost, what is it that im supposed to calculate?

I think the word you want is "quotient", not "quota". And you mean the derivative "at" zero (though prepositions are highly variable, so maybe "in zero" is normal in your culture or language).

But it isn't clear what "decide the quot[ient]" means. The "difference quotient" is the quotient before taking the limit, so it doesn't sound like you are supposed to find the derivative.

Can you quote the question for us in the original language, so we can decide what it means?