isi-entrance 2009 Q6

isi-entrance · India · solved Taylor series Extract derivative values from a given series
Let $\log x = g(x) = x f(x)$. Find $f^{(n)}(1)$, the $n$-th derivative of $f$ evaluated at $x = 1$.
$$\log x = g(x) = xf(x)$$ $$(-1)^2 x^{-1} = g'(x) = xf'(x) + f(x)$$ $$(-1)^3 x^{-2} 1! = g''(x) = xf''(x) + 2f'(x)$$ $$(-1)^4 x^{-3} 2! = g'''(x) = xf'''(x) + 3f''(x)$$ $$(-1)^{n+1}(n-1)! x^{-n} = g^{(n)}(x) = xf^{(n)}(x) + nf^{(n-1)}(x)$$ Evaluating at $x=1$: $$(-1)^{n+1}(n-1)! = f^{(n)}(1) + nf^{(n-1)}(1)$$ From induction: $$f^{(n)}(1) = (-1)^{n+1}(n-1)! - nf^{(n-1)}(1) = (-1)^{n+1}(n)!/n$$
Let $\log x = g(x) = x f(x)$. Find $f^{(n)}(1)$, the $n$-th derivative of $f$ evaluated at $x = 1$.