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?
"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.
I got it, thanks :)
Great!
To find the derivative of a function, you should use the limit definition of the derivative:
f'(a) = lim h->0 [f(a+h) - f(a)] / h
For the given function f(x) = (1 + 2x) / (1 + x), let's find f'(a) using the limit definition.
Substitute the function into the definition:
f'(a) = lim h->0 [(1 + 2(a + h))/(1 + (a + h))] - [(1 + 2a)/(1 + a)] / h
Now, we'll simplify the expression:
f'(a) = lim h->0 [(1 + 2a + 2h)/(1 + a + h) - (1 + 2a)/(1 + a)] / h
Combine the fractions with common denominators:
f'(a) = lim h->0 [(1 + 2a + 2h - 1 - 2a)/(1 + a + h)(1 + a)] / h
Cancel out 1 + 2a - 1 - 2a:
f'(a) = lim h->0 [2h / (1 + a + h)(1 + a)] / h
Now, let's simplify further by factoring out h from the numerator:
f'(a) = lim h->0 [2h / h * (1 + a + h)(1 + a)]
Cancel out h in the numerator and denominator:
f'(a) = lim h->0 2 / (1 + a + h)(1 + a)
Now, evaluate the limit as h approaches 0:
f'(a) = 2 / (1 + a)(1 + a) = 2 / (1 + a)^2
So, f'(a) = 2 / (1 + a)^2.