Question

If $y = \cos x,{y_n} = \dfrac{{{d^n}\left( {\cos x} \right)}}{{d{x^n}}}$, then $\left| {\begin{array}{*{20}{c}}
  {{y_4}}&{{y_5}}&{{y_6}} \\
  {{y_7}}&{{y_8}}&{{y_9}} \\
  {{y_{10}}}&{{y_{11}}}&{{y_{12}}}
\end{array}} \right|$=....
$\left( a \right)0$
$\left( b \right) - \cos x$
$\left( c \right)\cos x$
$\left( d \right)\sin x$

Answer 1

Hint: In this particular question use the concept that the differentiation of cos x is – sin x and the differentiation of sin x is cos x, later on use the concept of expansion of determinant so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given equation
$y = \cos x$.............. (1)
${y_n} = \dfrac{{{d^n}\left( {\cos x} \right)}}{{d{x^n}}}$
Now as we know that the differentiation of cos x is – sin x, and the differentiation of sin x is cos x so differentiate equation (1) w.r.t x we have,
$ \Rightarrow {y_1} = \dfrac{{d\left( {\cos x} \right)}}{{dx}} = - \sin x$
Now again differentiate w.r.t x we have,
$ \Rightarrow {y_2} = - \dfrac{d}{{dx}}\sin x = - \cos x$
Now again differentiate w.r.t x we have,
$ \Rightarrow {y_3} = - \dfrac{d}{{dx}}\cos x = \sin x$
Now again differentiate w.r.t x we have,
$ \Rightarrow {y_4} = \dfrac{d}{{dx}}\sin x = \cos x$
So as we see that ${y_4}$ is the same as y, so the value will be repeated.
$ \Rightarrow {y_5} = y = {y_9} - \sin x$
$ \Rightarrow {y_6} = {y_2} = {y_{10}} = - \cos x$
$ \Rightarrow {y_7} = {y_3} = {y_{11}} = \sin x$
$ \Rightarrow {y_8} = {y_4} = {y_{12}} = \cos x$
Now substitute these values in the given determinant we have,
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {\cos x}&{ - \sin x}&{ - \cos x} \\
  {\sin x}&{\cos x}&{ - \sin x} \\
  { - \cos x}&{\sin x}&{\cos x}
\end{array}} \right|\]
Now expand this determinant we have,
\[ \Rightarrow \cos x\left| {\begin{array}{*{20}{c}}
  {\cos x}&{ - \sin x} \\
  {\sin x}&{\cos x}
\end{array}} \right| - \left( { - \sin x} \right)\left| {\begin{array}{*{20}{c}}
  {\sin x}&{ - \sin x} \\
  { - \cos x}&{\cos x}
\end{array}} \right| - \cos x\left| {\begin{array}{*{20}{c}}
  {\sin x}&{\cos x} \\
  { - \cos x}&{\sin x}
\end{array}} \right|\]
Now expand the mini determinant we have,
\[ \Rightarrow \cos x\left( {{{\cos }^2}x - \left( { - {{\sin }^2}x} \right)} \right) - \left( { - \sin x} \right)\left( {\sin x\cos x - \left( { - \sin x} \right)\left( { - \cos x} \right)} \right) - \cos x\left( {{{\sin }^2}x - \left( { - {{\cos }^2}x} \right)} \right)\]
Now simplify it we have,
\[ \Rightarrow \cos x\left( {{{\cos }^2}x + {{\sin }^2}x} \right) + \sin x\left( {\sin x\cos x - \sin x\cos x} \right) - \cos x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\]
\[ \Rightarrow \cos x\left( {{{\cos }^2}x + {{\sin }^2}x} \right) + 0 - \cos x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\]
\[ \Rightarrow \cos x\left( {{{\cos }^2}x + {{\sin }^2}x} \right) - \cos x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\]
\[ \Rightarrow 0\]
Hence option (a) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic differentiation of sin x as well as cos x which is stated above, and always recall how to expand the determinant so expand it as above and simplify we will get the required answer.