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}}}{{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.