Confused: using Taylor series to find derivative TL;DR: read bolded parts
Lets say I have f(x) = sin(x^2) and I want the f''''''(0) (6th derivative). Using taylor series, this is really simple. We plug in x^2 into the taylor polynomial of sin(x), and get this:

Then the 6th derivative is 1/3! * 6! = 120.
I am confused because taylor series seems really unrelated; there should be an equally easy way to do this just with derivatives and chain rule (no detour to taylor series). But when I bash it out, I don't get a simple solution.

(120 on the last line, typo)
Why does taylor series come up in finding derivatives?
 A: There is (a-priori) no reason to believe, that calculating the sixth derivative of a function
will be easy.
In fact, if you were to fully write down the 6th derivative, this would be a really complicated expression.
The reason that the trick with the Taylor polynomial works so well here
is because $\sin(x)$ has a nice and well-known Taylor polynomial at the point $x=0$.
And for many functions, calculating the Taylor polynomial is also some work,
because it requires calculating derivatives.
And overall, Taylor polynomials already provide a lot of useful information about derivatives,
which is one of the reasons why mathematicians like Taylor polynomials.
A: Ok, perhaps there is another  way!
$$ f= \sin(x^2) $$
Rewrite this as:
$$ f= \sin(u)$$
where $ u =x^2$, with  $ \frac{du}{dx} = 2x $ , $ \frac{d^2 u}{dx^2} = 2 $ and$ \frac{d^3 u}{dx^3} = 0$
then:
$$ \frac{df}{dx}= g_1 = \cos u \frac{du}{dx}$$
At this point, we can apply the leibniz rule for derivatives (I wrote an article on this ),
$$ \frac{d^5 g_1}{dx^5} = \sum_{i=0}^5 \binom{5}{i} \frac{d^{i+1} u}{dx^{i+1}} \frac{d^{5-i} \cos u}{d^{5-i} x}$$
Now, the important thing to notice that the sum is zero except for $ i=1$ at x=0, (can you figure out why?). Hence,
$$ \frac{d^5 g_1}{dx^5} = f= \binom{5}{1} (2) \frac{d^4}{dx^4} \cos u |_{0}$$
Now rinse and repeat:
$$ g_2 = \frac{d^4}{dx^4} \cos u  =-\frac{d^3}{dx^3} ( \frac{du}{dx} \sin u ) =- \sum_{i=0}^3 \binom{3}{i} \frac{d^{u+i} u}{dx^{u+i}} \frac{d^{3-i} \sin u }{d^{3-i}x}=-\binom{3}{1}(2) \frac{d^2}{dx^2} \sin(u)$$
And again,
$$ g_3 = -\frac{d^2}{dx^2} \sin(u) = -\frac{d}{dx} ( \cos(u) \frac{du}{dx}) = 2\cos(u) \binom{1}{1}+ \text{stuff going to zero}$$
Piecing everything together,
$$ f= \binom{5}{1} 2 \binom{3}{1} 2 \binom{1}{1} 2 \cos(u)$$
Evaluating 'officially' at $x=0$
$$ f(0)=120$$
