A) 2
B) 4
C) 0
D) 8
Correct Answer: C
Solution :
\[\frac{d''[f(x)]}{d{{x}^{n}}}\,\,i.e.\,\,{{f}^{n}}(x)\] ? (I) [since elements in second and third rows are constants] Let \[y={{x}^{n}}\], then \[{{y}_{1}}=\frac{dy}{dx}=n{{x}^{n-1}},\,\,{{y}_{2}}=\frac{{{d}^{2}}y}{d{{x}^{2}}}=n(n-1){{x}^{n-2}}\], ...II \[{{y}_{n}}=n(n-1)(n-2)....1.{{x}^{n-n}}=n!\] Again let \[y=sin\,x\], then \[{{y}_{1}}\ =cos\,x=sin\,\left( \frac{\pi }{2}+x \right)\] \[{{y}_{2}}\ =\,\,cos\,\left( \frac{\pi }{2}+x \right)=sin\,\left\{ \frac{\pi }{2}+\left( \frac{\pi }{2}+x \right) \right\}=\sin \left( 2.\frac{\pi }{2}+x \right)\]Proceeding in this way, we get \[{{\operatorname{y}}_{n}}=\,\,sin\,\left( n.\frac{\pi }{2}+x \right)\] Similarly, if \[\operatorname{y} =cos\,x\], then \[{{\operatorname{y}}_{1}} =-\sin \,x=cos\left( \frac{\pi }{2}+x \right)\] \[{{\operatorname{y}}_{2}}\ =\,\,-\sin \,\left( \frac{\pi }{2}+x \right)=\cos \,\left\{ \frac{\pi }{2}+\left( \frac{\pi }{2}+x \right) \right\}\] \[=\,\,\cos \left( 2.\frac{\pi }{2}+x \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \,\,{{y}_{n}}=\cos \left( n\,.\frac{\pi }{2}+x \right)\] Thus, \[\frac{{{d}^{n}}}{d{{x}^{n}}}({{x}^{n}})=n!,\,\,\frac{{{d}^{n}}}{d{{x}^{n}}}(sin\,x)=sin\left( x+\frac{n\pi }{2} \right)\] and \[\frac{{{d}^{n}}}{d{{x}^{n}}}(\cos \,x)=\cos \,\left( x+\frac{n\pi }{2} \right)\] From (1), we have \[[\because \,\,\,{{R}_{1}}\,\,and\,\,{{R}_{2}}\,\,are\,\,identical]\]You need to login to perform this action.
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