Calculate-d-n-dx-n-sinx-d-n-dx-n-lnx-

Question Number 108350 by Ar Brandon last updated on 16/Aug/20

Calculate \frac{d^{n}}{{dx}^{n}} (sinx), \frac{d^{n}}{{dx}^{n}} (lnx)

Answered by Dwaipayan Shikari last updated on 16/Aug/20

f (x) = l o g x

f^{'} (x) = \frac{1}{x}

f^{″} (x) = - \frac{1}{x^{2}}

f^{‴} (x) = \frac{2}{x^{3}}

f^{(i v)} = - \frac{6}{x^{4}}

f^{(v)} = \frac{24}{x^{5}}

\dots f^{(n)} = {(- 1)}^{n + 1} . \frac{(n - 1)!}{x^{n}}

Answered by ajfour last updated on 16/Aug/20

y = \sin x

\frac{d y}{d x} = y_{1} = \sin (\frac{π}{2} + x)

y_{2} = \sin [2 (\frac{π}{2}) + x]

\dots .

y_{n} = \sin [n (\frac{π}{2}) + x]

Answered by mathmax by abdo last updated on 17/Aug/20

\frac{d^{n}}{{dx}^{n}} (sinx) = \sin (x + \frac{n π}{2}) we can prove this by recurrence

\frac{d}{dx} (lnx) = \frac{1}{x} \Rightarrow \frac{d^{n}}{{dx}^{n}} (lnx) = {(\frac{1}{x})}^{(n - 1)} = \frac{{(- 1)}^{n - 1} (n - 1)!}{x^{n}}

for n ⩾ 1