Question

If \[f(x)=\left| \begin{matrix}
   {{x}^{n}} & n! & 2 \\
   \cos x & \cos \dfrac{n\pi }{2} & 4 \\
   \sin x & \sin \dfrac{n\pi }{2} & 8 \\
\end{matrix} \right|\] , then find the value of \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}{{\left[ f(x) \right]}_{x=0}}\] .

Answer 1

Hint:The value of \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}\left| \begin{matrix}
   f(x) & {{a}_{1}} & {{a}_{2}} \\
   g(x) & {{b}_{1}} & {{b}_{2}} \\
   h(x) & {{c}_{1}} & {{c}_{2}} \\
\end{matrix} \right|\] where \[{{a}_{1}},{{a}_{2}},{{b}_{1}},{{b}_{2}},{{c}_{1}}\] and \[{{c}_{2}}\] are constant and is equal to\[\dfrac{{{d}^{n}}}{d{{x}^{n}}}\left| \begin{matrix}
   \dfrac{{{d}^{n}}}{d{{x}^{n}}}f(x) & {{a}_{1}} & {{a}_{2}} \\
   \dfrac{{{d}^{n}}}{d{{x}^{n}}}g(x) & {{b}_{1}} & {{b}_{2}} \\
   \dfrac{{{d}^{n}}}{d{{x}^{n}}}h(x) & {{c}_{1}} & {{c}_{2}} \\
\end{matrix} \right|\] .

Complete step-by-step answer:
We are given the determinant\[f(x)=\left| \begin{matrix}
   {{x}^{n}} & n! & 2 \\
   \cos x & \cos \dfrac{n\pi }{2} & 4 \\
   \sin x & \sin \dfrac{n\pi }{2} & 8 \\
\end{matrix} \right|\] .
We need to find the value of \[{{n}^{th}}\] derivative of the function \[f(x)\] with respect to \[x\] at \[x=0\], i.e. we need to find the value of \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}{{\left[ f(x) \right]}_{x=0}}\] .
First , we will find the value of \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}\left[ f(x) \right]\] , i.e. the value of \[{{n}^{th}}\] derivative of the function \[f(x)\] with respect to \[x\] . To find the value of \[{{n}^{th}}\] derivative of the function \[f(x)\] with respect to \[x\] , we will differentiate the function \[f(x)\] \[n\] times with respect to \[x\] .
On differentiating the function \[n\] times with respect to \[x\], we will get,
\[\dfrac{{{d}^{n}}}{d{{x}^{n}}}f(x)=\left| \begin{matrix}
   \dfrac{{{d}^{n}}}{d{{x}^{n}}}{{x}^{n}} & n! & 2 \\
   \dfrac{{{d}^{n}}}{d{{x}^{n}}}\cos x & \cos \dfrac{n\pi }{2} & 4 \\
   \dfrac{{{d}^{n}}}{d{{x}^{n}}}\sin x & \sin \dfrac{n\pi }{2} & 8 \\
\end{matrix} \right|\]
Now, we need to find the values of \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}{{x}^{n}}\] , \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}\cos x\] and \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}\sin x\] .
First , we will find the value of \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}{{x}^{n}}\] .
We know , \[\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}};\dfrac{{{d}^{2}}}{d{{x}^{2}}}{{x}^{n}}=(n)(n-1){{x}^{n-2}};\dfrac{{{d}^{3}}}{d{{x}^{3}}}{{x}^{n}}=(n)(n-1)(n-2){{x}^{n-3}}\] and so on.
On following the pattern , we can conclude: \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}{{x}^{n}}=(n)(n-1)(n-2)(n-3).....1=n!\] .
Now , we will find the value of \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}\cos x\] .
We know,
     \[\dfrac{d}{dx}\cos x=-\sin x=\cos \left( \dfrac{\pi }{2}+x \right);\dfrac{{{d}^{2}}}{d{{x}^{2}}}\cos x=-\cos x=\cos \left( \dfrac{2\pi }{2}+x \right);\dfrac{{{d}^{3}}}{d{{x}^{3}}}\cos x=\sin x=\cos \left( \dfrac{3\pi }{2}+x \right)\] and so on.
On following the pattern, we can conclude: \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}\cos x=\cos \left( \dfrac{n\pi }{2}+x \right)\] .
Now , we will find the value of \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}\sin x\].
We know ,
\[\dfrac{d}{dx}\sin x=\cos x=\sin \left( \dfrac{\pi }{2}+x \right);\dfrac{{{d}^{2}}}{d{{x}^{2}}}\sin x=-\sin x=\sin \left( \dfrac{2\pi }{2}+x \right);\dfrac{{{d}^{3}}}{d{{x}^{3}}}\sin x=-\cos x=\sin \left( \dfrac{3\pi }{2}+x \right)\] and so on.
On following the pattern, we can conclude: \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}\sin x=\sin \left( \dfrac{n\pi }{2}+x \right)\] .
Substituting these values in \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}f(x)\] , we get,
\[\dfrac{{{d}^{n}}}{d{{x}^{n}}}\left[ f(x) \right]=\left| \begin{matrix}
   n! & n! & 2 \\
   \cos \left( \dfrac{n\pi }{2}+x \right) & \cos \dfrac{n\pi }{2} & 4 \\
   \sin \left( \dfrac{n\pi }{2}+x \right) & \sin \dfrac{n\pi }{2} & 8 \\
\end{matrix} \right|\] .
Now , to find the value of \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}{{\left[ f(x) \right]}_{x=0}}\] , we will substitute \[x=0\] in \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}f(x)\] .
So, \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}{{\left[ f(x) \right]}_{x=0}}=\left| \begin{matrix}
   n! & n! & 2 \\
   \cos \left( \dfrac{n\pi }{2} \right) & \cos \dfrac{n\pi }{2} & 4 \\
   \sin \left( \dfrac{n\pi }{2} \right) & \sin \dfrac{n\pi }{2} & 8 \\
\end{matrix} \right|\]
Clearly , we can see that the two columns in the determinant are the same. Hence , the value of the determinant is zero.
So, \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}{{\left[ f(x) \right]}_{x=0}}=0\].

Note: While finding the value of \[{{n}^{th}}\] derivative of \[\sin x\] and \[\cos x\] with respect to \[x\] , be careful of the sign convention. Students generally make mistakes in the sign and end up getting a wrong answer.