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.