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I'm working a practice problem for Calculus 2 and I can't figure out a way to proceed, I'm stuck.

The function is: $$ h(t)=\arcsin(t^2) $$ The requested value to calculate with the differential is: $$ \arcsin(0.6) $$

I understand the theory and have already done similar problems but those involved square roots, polynomial expressions and others . They were easier.

The differential for h is $dh=h'(t)·dt$, so to calculate it one would use: $$ h(t+∆t)=h(t)+h'(t)·∆t $$ But I tried making $x=0 $, $x=-1$, $x=1$ and it didn't work because it resulted in an indetermination or a wrong value. I can't think of any other value to plug in the inverse sine function so the result is known or exact.

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HINT
Set $t=1/2$ and $\Delta t=0.1$ and remember that $\sin (\frac{\pi}{6})=1/2$

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    Right! But in the function the variable is squared and the value changes to 1/4, how can I overcome that? – 2017-01-07
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    Is the task to compute $h(0.6)$ or $\arcsin(0.6)$? – 2017-01-07
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    It's arcsin(0.6) but in the given function the variable is squared – 2017-01-07
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    You are being asked to calculate $\arcsin (0.6)$. It seems to me the given $h(t)$ is irrelevant for this problem. – 2017-01-07
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    Maybe there's a stability or quickness of convergence issue that makes $h$ a stronger choice than just plain $\arcsin$(?) – 2017-01-07
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    Yeah, for me it doesn't make sense that the problem gives you a function that is squared when you can use the plain one that isn't and get the same result. Thank you all for commenting! – 2017-01-07