help with derivative problem!?

Question

it says find the derivative.. and the equation to find it as u know is Lim h-->0 f(x+h) - f(x) / h


1. r(s) = sq.rt(2s+1)


thanks

Answer 1

1) Find r(s+h).
r(s+h) = sqrt(2(s+h)+1) = sqrt(2s+2h+1)

2) Subtract r(s)
r(s+h) - r(s) = sqrt(2s+2h+1) - sqrt(2s+1)

3) Divide by h
[r(s+h) -r(s)] / h
= [sqrt(2s+2h+1) - sqrt(2s+1)] / h

4) Simplify.
Multiply the top and bottom by sqrt(2s+2h+1) + sqrt(2s+1)

[sqrt(2s+2h+1) - sqrt(2s+1)][sqrt(2s+2h+1) + sqrt(2s+1)] / [hsqrt(2s+2h+1) + sqrt(2s+1)]
= (2s+2h+1-(2s+1)] / [hsqrt(2s+2h+1) + sqrt(2s+1)]
= 2h / [hsqrt(2s+2h+1) + sqrt(2s+1)]
Cancel the h's
= 2 / [sqrt(2s+2h+1) + sqrt(2s+1)]

5) Take the limit as h goes to zero.
= 2 / [sqrt(2s+2(0)+1) +sqrt(2s+1)]
= 2 / [sqrt(2s+1) + sqrt(2s+1)]
= 2/ [2sqrt(2s+1)]
= 1/sqrt(2s+1)

Answer 2

You don't want to use the definition of derivative directly. You want to use the chain rule:

if h(x) = g(f(x)) then

dh/dx = (dg/du)(df/dx)

in this case:

f(x) = 2x+1
g(u) = sqrt(u)

so:

1. compute df/dx (also known as f'(x))

2. compute dg/du (aka g')

3. substitute for u (u = f(x) = 2x+1) in the result you got for g' (the original computation of g' was in terms of u but the result needs to be in terms of x)

4. then multiply g' and f' to get the desired result.