# calculus II

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This image has been Flagged as inappropriate Click to unflag `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