`f'(x)=-1/(3-x)^2*(-1)=1/(3-x)^2`
`f''(x)=(1*2)/(3-x)^3`
`f'''(x)=(1*2*3)/(3-x)^4`
`f^((4))(x)=(1*2*3*4)/(3-x)^5`
Now we can see the pattern so we can write `f^((n))(x).`
The denominator is obviously `(3-x)^(n+1)` (exponent is by 1 greater then the degree of derivation). Numerator is `n!,` hence
`f^((n))(x)=(n!)/(3-x)^(n+1)` <-- n-th derivative` `
You can prove this formally by using mathematical induction.
2)
Formula for Taylor series is `f(x)=sum_(n=0)^oo (f^((n))(x))/(n!)(x-a)^n` . Since in your case `a=4` you have
`f(x)=sum_(n=0)^oo ((n!)/(3-4)^(n+1))/(n!)(x-4)^n=sum_(n=0)^oo(-1)^(n+1)(x-4)^n` <-- Your Taylor series
3)
Again we first find several derivatives.
`g(x)=sinh2x`
`g'(x)=2cosh2x`
`g''(x)=2^2sinh2x`
`g'''(x)=2^3cosh2x`
`g^(4)(x)=2^4sinh2x`
Here we are looking for Maclaurin series which is actually Taylor series about 0. Since `sinh0=0` and `cosh0=1` we will only be interested for odd derivatives (even derivatives contain `sinh0` which is equal to 0) and (2n+1)-th derivative is equal to `2^(2n+1).` Thus we have
`sinh2x=sum_(n=0)^oo(2^(2x+1)x^(2n+1))/((2n+1)!)` <-- Your Maclaurin series