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  1. $f(t)=\dfrac{\ln t}{\sqrt t}$

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?

Jim
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  • Well, substitution always makes it easier. In the second problem, putting $x=\sin t$ would be helpful. In the first one, won't be much helpful but, try putting $t=e^y$ – Shubham May 22 '14 at 19:11
  • I've never been taught the substitution methods for differentiation, so for now I'd like to stick to methods I already know. Any other methods? – Jim May 22 '14 at 19:23
  • Have you been taught the chain rule? – Shubham May 22 '14 at 19:27
  • Yes - I have, but I'm having a serious brain lag here. – Jim May 22 '14 at 19:32
  • There are no 'methods' for substitution. I mean, you've got to see how possibly you could simplify it. For the second problem, putting $x=sin t$ gives $arccos(1-sin^2 t)=arccos(cos 2t)=2t=2 arcsin x$, you see what I mean? – Shubham May 22 '14 at 19:39

2 Answers2

1

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.

Mike
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1

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 $x=0$.

Barry Cipra
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  • But surely using the quotient rule here is still fine? Low d high minus high d low over low squared... Not sure why but I have a preference for it. I'll try the product rule for this one then, thanks. – Jim May 22 '14 at 19:58
  • @Jim, yes, the quotient rule is OK to use, you just wind up with more clean-up to do. Also, if you compare what you got for $f'(t)$ with what I got, you'll see that one of us made a mistake, because I got a $t^{-3/2}$ and you got (in effect) a $t^{-5/2}$. Which one of us do you think is wrong? – Barry Cipra May 22 '14 at 20:03
  • Ah, that was me being silly, I squared $t$ instead of $\sqrt t$ – Jim May 22 '14 at 20:28
  • I tried it and got $t^{-5/2}(-3/2(1-0.5 \ln t)+0.5)$... correct? – Jim May 22 '14 at 20:40
  • @Jim, I get a $-0.5$ at the end, or $(3\ln t-8)/(4t^{5/2})$ in all. – Barry Cipra May 22 '14 at 20:46