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Question Number 146899 by mathmax by abdo last updated on 16/Jul/21
f(x)=sin^5 x   calculate f^((5)) ((π/2))
$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}^{\mathrm{5}} \mathrm{x}\:\:\:\mathrm{calculate}\:\mathrm{f}^{\left(\mathrm{5}\right)} \left(\frac{\pi}{\mathrm{2}}\right) \\ $$
Answered by Olaf_Thorendsen last updated on 17/Jul/21
f(x) = sin^5 x  f(x) = (((e^(ix) −e^(ix) )/(2i)))^5   f(x) = (1/(16×2i))(e^(5ix) −5e^(3ix) +10e^(ix) −10e^(−ix) +5e^(−3ix) −e^(−5ix) )  f(x) = ((sin5x−5sin3x+10sinx)/(16))  f^((5)) (x) = ((5^5 cos5x−5.3^5 cos3x+10cosx)/(16))  f^((5)) ((π/2)) = 0
$${f}\left({x}\right)\:=\:\mathrm{sin}^{\mathrm{5}} {x} \\ $$$${f}\left({x}\right)\:=\:\left(\frac{{e}^{{ix}} −{e}^{{ix}} }{\mathrm{2}{i}}\right)^{\mathrm{5}} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{16}×\mathrm{2}{i}}\left({e}^{\mathrm{5}{ix}} −\mathrm{5}{e}^{\mathrm{3}{ix}} +\mathrm{10}{e}^{{ix}} −\mathrm{10}{e}^{−{ix}} +\mathrm{5}{e}^{−\mathrm{3}{ix}} −{e}^{−\mathrm{5}{ix}} \right) \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{sin5}{x}−\mathrm{5sin3}{x}+\mathrm{10sin}{x}}{\mathrm{16}} \\ $$$${f}^{\left(\mathrm{5}\right)} \left({x}\right)\:=\:\frac{\mathrm{5}^{\mathrm{5}} \mathrm{cos5}{x}−\mathrm{5}.\mathrm{3}^{\mathrm{5}} \mathrm{cos3}{x}+\mathrm{10cos}{x}}{\mathrm{16}} \\ $$$${f}^{\left(\mathrm{5}\right)} \left(\frac{\pi}{\mathrm{2}}\right)\:=\:\mathrm{0} \\ $$

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