Question

Q1Find the nth derivative of ex log

Answer 1

Step-by-step explanation:

SOLUTION

TO DETERMINE

The n th derivative of

$\sf {e}^{x} \log x$

EVALUATION

Here the given function is

$\sf f(x) = {e}^{x} \log x$

$\sf \: Let \: \: \: u = {e}^{x} \: \: and \: \: v = \log x$

Then f(x) = uv

Now

$\sf \:u_n = {e}^{x} \: \: for \: all \: n$

Again

$\displaystyle \sf \: v_n = \frac{ {( - 1)}^{n - 1} \: (n - 1)!}{ {x}^{n} }$

Now f(x) = uv

Differentiating both sides n times with respect to x using Leibnitz theorem we get

$</p><p>\displaystyle \sf \: {f}^{n}(x) = (uv) _n = \displaystyle \sf\sum\limits_{r = 0}^{n} \: {}^{n} C_ru_{n - r}v_r$

$\displaystyle \sf \implies {f}^{n}(x) = \displaystyle \sf\sum\limits_{r = 0}^{n} \: {}^{n} C_ru_{n - r}v_r$

$\displaystyle \sf \implies {f}^{n}(x) = {e}^{x} \log x + \displaystyle \sf\sum\limits_{r=1}^{n} \: {}^{n} C_r \: {e}^{x} \: \frac{ {( - 1)}^{r - 1} \: (r - 1)!}{ {x}^{r} }$

$\displaystyle \sf \implies {f}^{n}(x) = {e}^{x} \log x + {e}^{x} \: \displaystyle \sf\sum\limits_{i=1}^{n} \: {}^{n} C_r \: \frac{ {( - 1)}^{r - 1} \: (r - 1)!}{ {x}^{r} }$

$</p><p>\displaystyle \sf \implies {f}^{n}(x) = {e}^{x} \bigg \{ \log x + n {x}^{ - 1} - {}^{n} C_2 {x}^{ - 2} + .. \: .. \bigg \}$

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