Let $f(x) = \ln(\arcsin(x^4) + \cos(x))$. Find $f^{(k)}(0)$ for $k=1,2,3,4$.
This is from my son's university calculus quiz which is already finished. He challenged me and I could not come up with a good solution. My son gave me a hint to make use of Taylor theorem though....
The original problem is $\arcsin(x^6)$ instead of $\arcsin(x^4)$ and to derive up to $k=6$, so directly differentiating 6 times with chain rule is not the expected answer.
Using your hint: express $f(x)$ as a Maclaurin series.
We have the well known series:
$$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}+ \frac{x^8}{8!} - \cdots$$
$$\arcsin x = x +\frac{x^3}{6} +\frac{3}{40}x^5 + \frac{5}{112}x^7 + \frac{35}{1152} x^9 + \cdots$$ Ref: https://mathworld.wolfram.com/InverseSine.html
$$\ln (1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} + \cdots$$
Carefully combining the above:
$$f(x) = \ln(\arcsin(x^4) + \cos(x))= \sum_{k=0}^\infty \frac{f^{(k)}}{k!} x^k =\sum_{k=0}^\infty a_k x^k= -\frac{x^2}{2}+\frac{11}{12} x^4 + \frac{43}{90}x^6 + O(x^8).$$
You can now read off the values from this series:
$$f^{(k)}(0) = k! a_k$$
So
$$\left\{f(0),f^\prime(0), f^{\prime \prime}(0), \cdots, f^{(6)}(0) \right\} = \left\{0,0,-1,0,22,0,344 \right\}$$
The strategy is to use substitutions into Maclaurin series. Here’s a compressed hint. (Note you can do all this in your head so long as you know all the basic series. If not, you might just need paper for the arcsine series.)
Write the argument of the logarithm as $1+(\arcsin(x^4)+\cos(x)-1)$ in order to pattern match with $\ln(1+x).$
Now use the Maclaurin expansions for arcsine (e.g. using Maclaurin expansion of $\arcsin x$ if it’s unknown in advance) and cosine to get $1+(x^4-x^2/2+x^4/24)$ up to order 4.
Finally, substitute into the first two terms of the Maclaurin for $\ln(1+x)$ to get $-x^2/2+11x^4/12$ up to order 4. (You need the first two terms because squaring the $x^2/2$ factor gives another order 4 contribution.)
From here you read off the derivatives.
Don't worry, this question just requires one to know a 'trick'
$$f(x) = \ln \left[ \arcsin(x^4) + \cos(x) \right]$$
We see that when $x=0$, the function inside log becomes a one, now we can manipulate the above as:
$$ f(x) = \ln \left[ 1+ \underbrace{[ \cos x + \arcsin(x^4) - 1 ]}\right]=\left[ \cos x + \arcsin(x^4) - 1 \right] - \frac{1}{2} \left[ \cos x + \arcsin(x^4) - 1 \right]^2...$$
Notice that the underbraced term is some small quantity in $h$ , near zero , so I applied the $\log(1+x)$ expansion. Now, consider the maclaurain of the quantity:
$$ g(x) = \cos(x) + \arcsin(x^4)-1$$
We can expand the above expansion from the standard expansions till the fourth order:
$$ g(x) = -\frac{x^2}{2!} + x^4( 1 + \frac{1}{4!}) +O(x^5)$$
Upon substitution into $f(x)$ we find,
$$ f(x) = - \left[ \frac{x^2}{2!} - x^4 (1 + \frac{1}{4!})\right]- \left[\frac12 \frac{x^4}{2!^2}\right] + O(x^6)$$
Now what you gotta do is, compare the coefficients of the powers in $x$ with the standard maclaurin coefficient for $f$, i.e: for $x^k$, it is $\frac{d^k f(x)}{dx^k}|_0 \frac{1}{k!}$
Just to give an alternative approach, it suffices to differentiate $f(x)$ once and then approximate things up to cubics in $x$ (since we're only taking three more derivatives). Using "$\approx$" to indicate we're ignoring powers of $x$ greater than $3$, we find
$$f'(x)={4x^3\left(1\over\sqrt{1-x^8}\right)-\sin x\over\arcsin(x^4)+\cos x}\approx{4x^3-\left(x-{1\over6}x^3 \right)\over1-{1\over2}x^2}\approx\left({25\over6}x^3-x\right)\left(1+{1\over2}x^2 \right)\approx-x+{11\over3}x^3$$
and thus $f''(x)\approx-1+11x^2$, $f'''(x)\approx22x$, and $f^{(4)}\approx22$, so we get $f'(0)=0$, $f''(0)=-1$, $f'''(0)=0$, and $f^{(4)}(0)=22$.
