find the nth e^x log x
Answers
Answer:
SOLUTION
TO DETERMINE
The n th derivative of
\sf {e}^{x} \log xe
x
logx
EVALUATION
Here the given function is
\sf f(x) = {e}^{x} \log xf(x)=e
x
logx
\sf \: Let \: \: \: u = {e}^{x} \: \: and \: \: v = \log xLetu=e
x
andv=logx
Then f(x) = uv
Now
\sf \:u_n = {e}^{x} \: \: for \: all \: nu
n
=e
x
foralln
Again
\displaystyle \sf \: v_n = \frac{ {( - 1)}^{n - 1} \: (n - 1)!}{ {x}^{n} }v
n
=
x
n
(−1)
n−1
(n−1)!
Now f(x) = uv
Differentiating both sides n times with respect to x using Leibnitz theorem we get
\displaystyle \sf \: {f}^{n}(x) = (uv) _n = \displaystyle \sf\sum\limits_{r = 0}^{n} \: {}^{n} C_ru_{n - r}v_rf
n
(x)=(uv)
n
=
r=0
∑
n
n
C
r
u
n−r
v
r
\displaystyle \sf \implies {f}^{n}(x) = \displaystyle \sf\sum\limits_{r = 0}^{n} \: {}^{n} C_ru_{n - r}v_r⟹f
n
(x)=
r=0
∑
n
n
C
r
u
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} }⟹f
n
(x)=e
x
logx+
r=1
∑
n
n
C
r
e
x
x
r
(−1)
r−1
(r−1)!
\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} }⟹f
n
(x)=e
x
logx+e
x
i=1
∑
n
n
C
r
x
r
(−1)
r−1
(r−1)!
\displaystyle \sf \implies {f}^{n}(x) = {e}^{x} \bigg \{ \log x + n {x}^{ - 1} - {}^{n} C_2 {x}^{ - 2} + .. \: .. \bigg \}⟹f
n
(x)=e
x
{logx+nx
−1
−
n
C
2
x
−2
+....}
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