derivative problem and MVT - need explanation

hey everyone i had 2 questions

firstly

If y = sqrt(x^2 + 1) , then the derivative of y^2 with respect to x^2 is

i'm not sure where to start and its a mc question so i tried to first take the derivative of x^2 and with respect to dy/dx but i don't think that is right

secondly

the mean value theorem states that f(x) has to be continuous at [a,b], a and b being the end points, and f(x) has to be differentiable at (a,b), again a and b being the end points. if this is true, are the end points included in the mean value theorem

thanks alot

Re: derivative problem and MVT - need explanation

Hello, ubhutto!

Quote:

$\displaystyle \text{Given: }\:y \:= \sqrt{x^2 + 1}$

$\displaystyle \text{Find the derivative of }y^2\text{ with respect to }x^2.$

Square both sides: .$\displaystyle y^2 \:=\:x^2+1$

Let $\displaystyle u = y^2,\;v = x^2$

Then we have: .$\displaystyle u \:=\:v+1$

Hence: .$\displaystyle \frac{du}{dv} \:=\:1$

Back-substitute: .$\displaystyle \frac{d(y^2)}{d(x^2)} \:=\:1$

Re: derivative problem and MVT - need explanation

With regards to your second question: yes they are included. The conditions ask for $\displaystyle f$ to be continuous on $\displaystyle [a,b]$. This is does **not** mean $\displaystyle f$ is continuous at $\displaystyle a$ and $\displaystyle b,$ (it* does* mean $\displaystyle f$ should be defined at those points) so we must allow $\displaystyle f$ to be non-differentiable at $\displaystyle a$ and $\displaystyle b,$ hence the open interval. For instance, the function:

$\displaystyle f(x) = \left\{ \begin{array}{l l} x^2 & \quad -1\leq x\leq 1\\0 & \quad \text{otherwise}\end{array} \right.$

is continuous on $\displaystyle [-1,1]$ but discontinuous ($\displaystyle \to$ not differentiable) at -1 and 1. The conditions mean that we may apply the theorem on $\displaystyle [-1,1]$ or any closed interval contained in $\displaystyle [-1,1]$.