calculus
posted by Janet .
Find the first and second derivative  simplify your answer.
y=x/4x+1
I solved the first derivative and got 1/(4x+1)^2
Not sure if I did the first derivative right and not sure how to do the second derivative.

The first derivative of 1/(4x+1)² is correct.
For the second derivative, you just have to differentiate the first derivative with respect to x.
You can use the chain rule, substituting u=4x+1, and differentiate 1/u² using the power rule.
Let
f'(x)=1/(4x+1)², and
u=4x+1
d(f'(x))/dx
=d(1/u²)du * du/dx
=2/u³ * (4)
=8/(4x+1)³ 
y = x/(4x +1)
dy/dx = [(4x+1)  4x]/(4x +1)^2
= 1/(4x +1)^2
OK so far
d^2y/dx^2 = 2(4x +1)^3 * 4
= 8/(4x+1)^3 
To solve the first step I used the equation
f'(x)=lim h>0 f(x+h)f(x)/h
Can I use this same formula to solve for the second derivative? 
Yes, if f"(x) replaces f'(x) on the left and f' replaces f on the right.

I plug in the numbers into the formula that I used, but I don't get the same answer.
Here's what I did, maybe you can tell me what I did wrong. (likely a dumb algebra mistake)
f"(x) = lim>0 f'(x+h)f'(x)/h
=(1/4(x+h)+1)^2  (1/4x+1)^2/h
=(1/16x^2+32xh+8x+8h+16h^2+1)(1/16x^2+8x+1)/h
=16x^2+8x+1(16x^2+32xh+16h^2+8x+8h+1)/
h(16x^2+32xh+16h^2+8x+8h+1)(16x^2+8x+1)
=32xh16h^28h /h(16x^2+32xh+16h^2+8x+8h+1)(16x^2+8x+1)
=8(4x+1)/(16x^2+8x+1)(16x^2+8x+1)
=8(4x+1)/(4x+1)^2
= 8/(4x+1)
I can't figure out where I went wrong.
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