If $y=x^{n-1}\ln (x)$, Then Prove That $xy_{n}=(n-1)!

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Learn more about Teams If $y=x^{n-1}\ln (x)$, then prove that $xy_{n}=(n-1)!$ [duplicate] Ask Question Asked 5 years ago Modified 5 years ago Viewed 6k times 0 $\begingroup$ This question already has answers here: If $f(x)=x^{n-1}\log x$, then the $n$-th derivative of $f$ is equals? (3 answers) Closed 5 years ago.

If $y=x^{n-1}\ln (x)$, then prove that $xy_{n}=(n-1)!$

My Attempt: $$y=x^{n-1}\ln (x)$$ Differentiating both sides with respect to $x$, $$y_{1}= x^{n-1} \dfrac {1}{x} + \ln (x)\cdot (n-1)x^{n-2}$$

P.S $y_{n}$ denotes the $n^{th}$ derivative of $y$.

• calculus
• derivatives

Share Cite Follow asked Nov 30, 2019 at 4:34 pi-πpi-π 7,55488 gold badges9898 silver badges188188 bronze badges $\endgroup$ 1

• 1 $\begingroup$ Your last three questions basically fall in the category of the same kind of question. What does really bother you about differentiation? That's a question worth of MSE. $\endgroup$ – Ennar Commented Nov 30, 2019 at 4:41

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Sorted by: Reset to default Highest score (default) Date modified (newest first) Date created (oldest first) 1 $\begingroup$

$$y=x^{n-1}\ln (x)$$ $$\implies y_1=(n-1)x^{n-2} \ln (x)+x^{n-2}$$ $$\implies xy_1-(n-1)y=x^{n-1}$$ Now differentiating $(n-1)$ times with the help of Leibnitz's Formula we have, $$xy_n+^{n-1}C_1y_{n-1}\cdot 1-(n-1)y_{n-1}=(n-1)!$$ $$xy_n+(n-1)y_{n-1}-(n-1)y_{n-1}=(n-1)!$$ $$\implies xy_n=(n-1)!$$

Leibnitz's Formula for the $n^{th}$ derivative of a product:

If $u$ and $v$ are functions of $x$, each possessing derivatives upto $n^{th}$h order, then the $n^{th}$ derivative of their product is given by

$$(uv)_n=u_nv+^nC_1 u_{n-1}v_1+^nC_2u_{n-2}v_2+\quad. . . \quad+^nC_ru_{n-r}v_r+\quad. . . \quad+ uv_n$$

Share Cite Follow edited Jun 12, 2020 at 10:38 CommunityBot 1 answered Nov 30, 2019 at 5:56 nmasantanmasanta 9,4522525 gold badges2828 silver badges5656 bronze badges $\endgroup$ Add a comment | 0 $\begingroup$

You can proceed by induction.

Note that if $y = x^n \ln x$ then $$y_1 = x^{n - 1} +n x^{n - 1} \ln x.$$ Taking derivative $n$ times and using the hypothesis give us $$y_{n + 1} = 0 + n\frac{(n - 1)!}{x}.$$ The conclusion follows.

Share Cite Follow answered Nov 30, 2019 at 4:50 AzlifAzlif 2,25777 silver badges2222 bronze badges $\endgroup$ Add a comment |

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