step 1 recall that the maclaurin series of a function f(x) is given by the following. f(x)=f(0)+f(0)x+(f(0)/2…

step 1 recall that the maclaurin series of a function f(x) is given by the following. f(x)=f(0)+f(0)x+(f(0)/2!)x^2+(f(0)/3!)x^3+(f^(4)(0)/4!)x^4+... to find the maclaurin series of the given function f(x), we first need to find all of its derivatives. find the first several derivatives. f(x)=2(1 - x)^(-2) f(x)=4(1 - x)^(-3) f(x)= f(x)= f^(4)(x)=

step 1 recall that the maclaurin series of a function f(x) is given by the following. f(x)=f(0)+f(0)x+(f(0)/2!)x^2+(f(0)/3!)x^3+(f^(4)(0)/4!)x^4+... to find the maclaurin series of the given function f(x), we first need to find all of its derivatives. find the first several derivatives. f(x)=2(1 - x)^(-2) f(x)=4(1 - x)^(-3) f(x)= f(x)= f^(4)(x)=

Answer

Explanation:

Step1: Differentiate $f'(x)$ to get $f''(x)$

Using the power - rule for differentiation $\frac{d}{dx}(u^n)=nu^{n - 1}u'$, where $u = 1 - x$ and $n=-3$. $f'(x)=4(1 - x)^{-3}$, so $f''(x)=4\times(- 3)(1 - x)^{-4}=-12(1 - x)^{-4}$

Step2: Differentiate $f''(x)$ to get $f'''(x)$

$f''(x)=-12(1 - x)^{-4}$, then $f'''(x)=-12\times(-4)(1 - x)^{-5}=48(1 - x)^{-5}$

Step3: Differentiate $f'''(x)$ to get $f^{(4)}(x)$

$f'''(x)=48(1 - x)^{-5}$, so $f^{(4)}(x)=48\times(-5)(1 - x)^{-6}=-240(1 - x)^{-6}$

Answer:

$f''(x)=-12(1 - x)^{-4}$, $f'''(x)=48(1 - x)^{-5}$, $f^{(4)}(x)=-240(1 - x)^{-6}$