Differentiation Question 402
Question: If $ y=\sin px $ and $ y _{n} $ is the nth derivative of y, then $ | \begin{matrix} y & y_1 & y_2 \\ y_3 & y_4 & y_5 \\ y_6 & y_7 & y_8 \\ \end{matrix} | $ is equal to
[AMU 2002]
Options:
A) 1
B) 0
C) - 1
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ D=| \begin{matrix} \sin px & p\cos px & -p^{2}\sin px \\ -p^{3}\cos px & p^{4}\sin px & p^{5}\cos px \\ -p^{6}\sin px & -p^{7}\cos px & p^{8}\sin px \\ \end{matrix} | $
$ =p^{9}| \begin{matrix} \sin px & p\cos px & -p^{2}\sin px \\ -\cos px & p\sin px & p^{2}\cos px \\ -\sin px & -p\cos px & p^{2}\sin px \\ \end{matrix} | $
$ =-p^{9}| \begin{matrix} \sin px & p\cos px & -p^{2}\sin px \\ \cos px & p\sin px & p^{2}\cos px \\ \sin px & p\cos px & -p^{2}\sin px \\ \end{matrix} |=0. $
BETA
