step 1 the taylor series formula is given. f(a)+f(a)(x - a)+f(a)/2!(x - a)^2 + f(a)/3!(x - a)^3 +…

step 1 the taylor series formula is given. f(a)+f(a)(x - a)+f(a)/2!(x - a)^2 + f(a)/3!(x - a)^3 + f^(4)(a)/4!(x - a)^4+... the function f(x)=7/x can also be written as f(x)=(7)1/x, which has derivatives f(x)=(7)-1/x^2, f(x)=(7)2/x^3, f(x)=(7)-6/x^4, and f^(4)(x)=(7)24/x^5. submit skip (you cannot come back) need help? read it submit answer

step 1 the taylor series formula is given. f(a)+f(a)(x - a)+f(a)/2!(x - a)^2 + f(a)/3!(x - a)^3 + f^(4)(a)/4!(x - a)^4+... the function f(x)=7/x can also be written as f(x)=(7)1/x, which has derivatives f(x)=(7)-1/x^2, f(x)=(7)2/x^3, f(x)=(7)-6/x^4, and f^(4)(x)=(7)24/x^5. submit skip (you cannot come back) need help? read it submit answer

Answer

Explanation:

Step1: Recall power - rule for differentiation

The power - rule states that if (y = x^n), then (y^\prime=nx^{n - 1}). For the function (f(x)=7x^{-1}), using the power - rule, (f^\prime(x)=7\times(- 1)x^{-1 - 1}=- \frac{7}{x^{2}}).

Step2: Differentiate (f^\prime(x))

Differentiating (f^\prime(x)=-7x^{-2}) again using the power - rule, (f^{\prime\prime}(x)=(-7)\times(-2)x^{-2 - 1}=\frac{14}{x^{3}}).

Step3: Differentiate (f^{\prime\prime}(x))

Differentiating (f^{\prime\prime}(x) = 14x^{-3}) using the power - rule, (f^{\prime\prime\prime}(x)=14\times(-3)x^{-3 - 1}=-\frac{42}{x^{4}}).

Step4: Differentiate (f^{\prime\prime\prime}(x))

Differentiating (f^{\prime\prime\prime}(x)=-42x^{-4}) using the power - rule, (f^{(4)}(x)=(-42)\times(-4)x^{-4 - 1}=\frac{168}{x^{5}}).

Answer:

(f^\prime(x)=-\frac{7}{x^{2}}, f^{\prime\prime}(x)=\frac{14}{x^{3}}, f^{\prime\prime\prime}(x)=-\frac{42}{x^{4}}, f^{(4)}(x)=\frac{168}{x^{5}})