Say I have function K(x,y). $$ K_x = \lim_{h \to 0} \frac{y[(x+h)(x+1)]^{2019} - y[x(x+h+1)]^{2019}}{h(x+1)^{2019}(x+h+1)^{2019}} $$ Find $K_{xxyy}$ (basically differentiate $x$ one more time then $y$ two times).
I tried solving it normally and noticed how big this was getting and finally realized that's not the correct way of doing it. I was thinking of perhaps factoring but didn't know how to start it. I noticed the first part of the numerator is like $f(x+h) - f(x)$, and the bottom is like $h$ but with two extra terms. Honestly I don't know how to work it and would appreciate some help. Thanks!
Denote $c = 2019$ and $G(x,h) = [(x+h)(x+1)]^c - [x(x+h+1)]^c$ we have $$K_x = \frac{y}{ (x+1)^c} \lim_{h \to 0} \frac1{(x+h+1)^c}\frac{ G(x,h)}{h}$$
Add and subtract two pairs of auxiliary terms to make three sets of $\Delta f = f(x+h) - f(x)$. $$\begin{alignat}{2} G(x,h) = [(x+h)(x+1)]^c & &{}- [x(x+1)]^c \\ &&{}+ [x(x+1)]^c &&{}- [(x+h)(x+h+1)]^c\\ && &&{}+ [(x+h)(x+h+1)]^c - [x(x+h+1)]^c \\ \end{alignat}$$ Or to emphasize the the structure of $f' = \lim \frac{\Delta f}h$ explicitly: $$\begin{align} \frac{ G(x,h) }h &= (x+1)^c \frac{(x+h)^c - x^c}h \\ & \hspace{24pt} + \frac{[(x+h)(x+1+h)]^c - [x(x+1)]^c}h \\ & \hspace{60pt} + (x+h+1)^c \frac{(x+h)^c - x^c}h \end{align}$$ Put back the common factor $\frac1{(x+h+1)^c}$ and take the limits of the three sets separately. The first set will combine with the 3rd set. $$\lim_{h \to 0} \frac1{(x+h+1)^c}\frac{ G(x,h) }h = \frac{\mathrm{d}x^c}{\mathrm{d}x} + \frac1{(x+1)}\frac{\mathrm{d}[x(x+1)]^c}{\mathrm{d}x} +\frac{\mathrm{d}x^c}{\mathrm{d}x}$$ Can you take it from here?
Note that $K_x$ is directly proportional to $y$ and so will $K_{xx}$ be. The double derivative with respect to $y$ will result zero $K_{xxyy} = 0$.
I have to assume that this is a typo in the question statement.
Anyway, the basic algebraic manipulation to formulate $f' = \lim \frac{\Delta f}h$ is demonstrated above. Please adapt accordingly after the typo involving $y$ is corrected.