Second derivative of $\frac{\ln t}{\sqrt t}$ and derivative of $\arccos(1-2x^2)$ 
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*$f(t)=\dfrac{\ln t}{\sqrt t}$


I'm stuck on the algebra of finding the second derivative.
For the first derivative, I got:
$f'(t)=\dfrac{t^{\frac{-1}{2}}(1-\frac{1}{2}\ln t)}{t^2}$
For the second derivative, I'm stuck on the algebra... If someone could differentiate this and show me the steps, I'd really appreciate it.
Also:
Differentiate $y=\arccos(1-2x^2)$ with respect to x, and simplify your answer.
So far I have:
$\dfrac{-4x}{\sqrt{4x^2-4x^4}}$
Am I on the right lines?
 A: It looks like you've made some mistakes.  As for the first one, that first derivative doesn't look right.  I've never really learned the rule for division as the product rule is much easier.
$$(t^{-1/2}\ln t)=t^{-3/2}-\frac12t^{-3/2}\ln t$$
Now repeat the process for the second derivative.  As for the second, it looks like you're close, but there are a couple problems.
$$d(\cos^{-1}t)=-\frac{dt}{\sqrt{1-t^2}}$$
If $t=1-2x^2$, we have
$$-\frac{d(1-2x^2)}{\sqrt{1-(1-2x^2)^2}}=-\frac{-4xdx}{\sqrt{1-(1-4x^2+4x^4)}}=\frac{4xdx}{\sqrt{4x^2-4x^4}}$$
From here, a little simplification can be done.  If you assume $x$ is positive, the denominator can be rewritten as $2x\sqrt{1-x^2}$ and you probably wouldn't be marked wrong if you rewrote it as such and cancelled the $x$'s.  The denominator is technically $2|x|\sqrt{1-x^2}$, but that doesn't cancel as nicely.
A: For the first problem, writing $f(t)=t^{-1/2}\ln t$ and then using the product rule might help:
$$f'(t)={-1\over2}t^{-3/2}\ln t+t^{-1/2}(1/t)=t^{-3/2}(1-{1\over2}\ln t)$$
See if you can now do the second derivative in the same way.
As for the arccos problem, you made a sign mistake.  The correct answer is
$${4x\over\sqrt{4x^2-4x^4}}$$
That is, the derivative of $\arccos t$ (with respect to $t$) is $-1/\sqrt{1-t^2}$.  You can simplify this to
$$2x\over|x|\sqrt{1-x^2}$$
but not to
$$2\over\sqrt{1-x^2}$$
unless the problem tells you that $x\gt0$.  In other words, as nice-looking as $\arccos(1-2x^2)$ seems to be, it does not have a derivative at $x=0$.
