# calculus

posted by on .

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.

• calculus - ,

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

• calculus - ,

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

• calculus - ,

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?

• calculus - ,

Yes, if f"(x) replaces f'(x) on the left and f' replaces f on the right.

• calculus - ,

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)
=-32xh-16h^2-8h /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.

#### Related Questions

More Related Questions

Post a New Question