# derivative problem and MVT - need explanation

• Jan 1st 2013, 09:22 AM
ubhutto
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
• Jan 1st 2013, 09:52 AM
Soroban
Re: derivative problem and MVT - need explanation
Hello, ubhutto!

Quote:

$\text{Given: }\:y \:= \sqrt{x^2 + 1}$
$\text{Find the derivative of }y^2\text{ with respect to }x^2.$

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

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

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

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

Back-substitute: . $\frac{d(y^2)}{d(x^2)} \:=\:1$
• Jan 1st 2013, 01:27 PM
LordoftheFlies
Re: derivative problem and MVT - need explanation
With regards to your second question: yes they are included. The conditions ask for $f$ to be continuous on $[a,b]$. This is does not mean $f$ is continuous at $a$ and $b,$ (it does mean $f$ should be defined at those points) so we must allow $f$ to be non-differentiable at $a$ and $b,$ hence the open interval. For instance, the function:

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

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