Q:

How could I find the $35$th derivative of this rational function?

I am struggling to find the $35$th derivative of $f(x)$ where
$$f(x)={1+x^{34}(x+3)^3(x+1)\over x^2(x+9)+27(x+1)}$$

A:

Hint:
Notice that your denominator is$$(x+3)^3$$
Then $f(x)$ can be rewritten as
$$f(x)=\frac{1+x^{34}(x+3)^3(x+1)}{x^2(x+9)+27(x+1)}=\frac{1+x^{35}(x+3)^3+x^{34}(x+3)^3}{(x+3)^3}$$
And then your entire expression can be reduced to $$x^{35}+x^{3
4}+\frac{1}{(x+3)^3}$$
I guess now it's easy to find $n$th derivatives of these terms easily
Note 1 :
$$\frac{\mathrm d^n}{\mathrm dx^n} x^n = n!$$
Don't see this unless you've solved it yourself

 $$\large\frac{\mathrm d^{35}}{\mathrm dx^{35}}f(x) =35!+0-\frac{37!}{2(x+3)^{38}}$$

A:

Hint: factorise the denominator, then simplify the expression for $f(x)$ to get. . . 

 $$f(x)=x^{35}+x^{34}+\dfrac1{(x+3)^3}\ .$$

