Thread: 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

Hello, ubhutto!

Square both sides: .

Let

Then we have: .

Hence: .

Back-substitute: .

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



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