00:01
In this question we need to use the limit definition to find the derivative of the function.
00:14
So for this we have the first function as y is equal to 2x square plus 1.
00:27
So the limit formula for evaluating the derivative is d .y by d x that will be equals to limit h tends to 0.
00:45
So f x plus h negative f of x divided by h.
00:55
So here we have the function f x is 2x square plus 1.
01:05
So x plus h will be that is replacing x by x plus h in this we get 2x plus h whole square plus 1.
01:20
Now we will put fx and f x plus h in this to evaluate the derivative using the limit formula.
01:30
So we have d y by d x will be equals to limit h tends to 0 to x plus h whole square positive 1 then we have negative sign 2x square plus 1 divided by h so here we get limit h tends to 0 now when we open it it can be written as 2x plus h whole square plus 1 negative 2 x square negative 1 divided by h so from this one this one will get cancelled and we get here we have limit h tends to 0 so it can be written as 2 x square plus h square negative 2 x h negative 2 x x squared divided by h, now it can be written as limit h -tens to 0, 2x square plus 2 -h -square negative 2x -h -h -negative 2x square divided by h.
03:04
So from this term, this term will get cancelled and it can be written as limit h tends to 0 on taking h and two common from the numerator we get to which here we have h negative x divided by h so sorry here instead of negative sign it will be a positive sign here also there is a positive sign here also a positive sign so from this h this h will get cancelled and now we put h is equal to 0 so we get 2 0 plus x which is equals to 2x...