00:01
We have today another question about working out a derivative algebraically.
00:05
So we need to work out this function, xq plus 5, at the point x equal 1.
00:13
So we know that the derivative of x is equal to the limit as h approaches 0 of f in the point 1 plus h minus f in the point 1 over h, which is equal to the limit.
00:35
As h approaches 0 of 1 plus h cube plus 5 minus 1 cubed plus 5 over h.
00:54
And this is equal to the limit as h approaches 0 of 1 plus 2h plus h squared times 1 plus h plus h plus 5 minus 1 plus 5.
01:18
Over h so we go and we go and simplify that one step further we get the limits as h approach zero of sorry of and i'm going to multiply now one with each of these and then h with each of them if you understand what i'm saying so so we have then 1 plus 2h plus h squared plus h squared plus h squared plus h cubed and then plus 5 minus 6 over h.
02:18
And if we go and look here, we see that we have 1 plus 5 minus 6, we have 2h plus h plus h, and we have 1 plus 5 plus h.
02:33
Have h squared plus 2h squared.
02:37
So then we go and work that out...