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