"lim h->0 [(1+2a+2h)(1+a+h)]-[(1+2a)(1+a)]/h =4a+2 "
Shouldn't it be
{[(1+2a+2h)(1+a)-(1+2a)(1+a+h)]/[(1+a+h)(1+a)]}/h
={[h]/[(1+a+h)(1+a)]}/h
=1/(1+a+h)(1+a)
I'm sure you can do the rest.
FInd f'(a) for the function, f(x)=(1+2x)/(1+x)
lim h->0 ][(1+2(a+h))/(1+a+h)]-[(1+2a)/(1+a)]]/h
I got rid of the denominator
lim h->0 [(1+2a+2h)(1+a+h)]-[(1+2a)(1+a)]/h =4a+2
What's wrong with this? Should I multiply the denominators with h as well?
3 answers
I got it, thanks :)
Great!
2 answers