Thread: Help needed finding the second derivative.

1. Help needed finding the second derivative.

Find the second derivative
$
f(x) = ((x-2)^3)/(x^2)
$

Attempt:

f'(x)/g(x) = (f'*g-g'*f)/(g^2)
$
f'(x) = ((3*(x-2)^2)*(x^2)-(2x)*((x-2)^3))/(x^4)
$

f(x)/g(x) = (f'*g-g'*f)/(g^2)
f(x)*d(x) = f'd + d'f

f''(x) = ((6(x-2)*x^2+2x*(3*(x-2)^2) -(2*((x-2)^3)+2x*(3(x-2)^2))*(x^4)-(4x^3)*((3*(x-2)^2)*(x^2)-(2x)*((x-2)^3))/(x^8)
*Sorry the math thing can't fit the size of this

Simplifying doesnt give me the right answer. The correct answer is $24(x-2)/(x^4)$.
How do i do this?

2. Originally Posted by Fallen186
Find the second derivative
$
f(x) = ((x-2)^3)/(x^2)
$

Attempt:

f'(x)/g(x) = (f'*g-g'*f)/(g^2)
$
f'(x) = ((3*(x-2)^2)*(x^2)-(2x)*((x-2)^3))/(x^4)
$

I do it again

f(x)/g(x) = (f'*g-g'*f)/(g^2)
f(x)*d(x) = f'd + d'f

f''(x) = ((6(x-2)*x^2+2x*(3*(x-2)^2) -(2*((x-2)^3)+2x*(3(x-2)^2))*(x^4)-(4x^3)*((3*(x-2)^2)*(x^2)-(2x)*((x-2)^3))/(x^8)
*Sorry the math thing can't fit the size of this

Simplifying doesnt give me the right answer. The correct answer is $24(x-2)/(x^4)$.
How do i do this?
This looksl like a prime candidate for logarthimic differenatiation.

Since $f(x)=\frac{(x-2)^3}{x^2}\implies \ln f(x)=\ln\left(\frac{(x-2)^3}{x^2}\right)=3\ln(x-2)-2\ln(x)$

Differentiating both sides gives

$\frac{f'(x)}{f(x)}=\frac{3}{x-2}-\frac{2}{x}\implies f'(x)=f(x)\left\{\frac{3}{x-2}-\frac{2}{x}\right\}$

But remember that $f(x)=\frac{(x-2)^3}{x^2}$

So $f'(x)=\frac{(x-2)^3}{x^2}\left\{\frac{3}{x-2}-\frac{2}{x}\right\}$

3. Originally Posted by Fallen186
Find the second derivative
$
f(x) = ((x-2)^3)/(x^2)
$

Attempt:

f'(x)/g(x) = (f'*g-g'*f)/(g^2)
$
f'(x) = ((3*(x-2)^2)*(x^2)-(2x)*((x-2)^3))/(x^4)
$
That is correct, but you should simplify it before differentiating it again.

$f'(x) = \frac{3(x-2)^2x^2-2x(x-2)^3}{x^4} = \ldots = \frac{(x-2)^2(x+4)}{x^3}$

Now differentiate, either by using the quotient rule again, or (preferably) by logarithmic differentiation as in Mathstud28's suggestion.