Find-the-nth-derivative-of-sin-2-2x-

Question Number 8996 by Basant007 last updated on 11/Nov/16

Find the nth derivative of \sin^{2} 2 x

Commented by FilupSmith last updated on 13/Nov/16

y = \sin^{2} (2 x)

u = \sin (2 x) \Rightarrow d u = 2 \cos (2 x) d x

d x = \frac{1}{2 \cos (2 x)} d u

\frac{d y}{d x} = \frac{d}{d x} (u^{2})

\frac{d y}{d x} = \frac{d}{\frac{1}{2 \cos (2 x)} d u} (u^{2})

\frac{d y}{d x} = 2 \cos (2 x) \frac{d}{d u} (u^{2})

\frac{d y}{d x} = 4 \cos (2 x) u

\frac{d y}{d x} = 4 \cos (2 x) \sin (2 x)

\cos (x) \sin (x) = \frac{1}{2} \sin (2 x)

\frac{d y}{d x} = 2 \sin (4 x)

\frac{d^{2} y}{{d x}^{2}} = 8 \cos (4 x)

\frac{d^{3} y}{{d x}^{3}} = - 32 \sin (4 x)

\frac{d^{4} y}{{d x}^{4}} = - 128 \cos (4 x)

I^{'} m not sure if there is a general form

for \frac{d^{n} y}{{d x}^{n}}

Answered by 123456 last updated on 14/Nov/16

y = \sin^{2} 2 x

\frac{d y}{d x} = 2 \sin 4 x

\frac{d^{2} y}{{d x}^{2}} = 8 \cos 4 x = 8 \sin (4 x + \frac{π}{2})

\frac{d^{3} y}{{d x}^{3}} = - 32 \sin 4 x = - 32 \sin (4 x + π)

\frac{d^{4} y}{{d x}^{4}} = - 128 \cos 4 x = - 128 \sin (4 x + \frac{3 π}{2})

\frac{d^{5} y}{{d x}^{5}} = 512 \sin 4 x

- - - - - - -

1 \to 2 = 2 \cdot 4^{0}; 0

2 \to 8 = 2 \cdot 4^{1} = 2^{3}; \frac{π}{2}

3 \to 32 = 2 \cdot 4^{2} = 2^{5}; π

4 \to 128 = 2 \cdot 4^{3} = 2^{7}; \frac{3 π}{2}

n \to 2 \cdot 4^{n - 1} = 2 \cdot 2^{2 n - 2} = 2^{2 n - 1}; \frac{(n - 1) π}{2}

\frac{d^{n} y}{{d x}^{n}} = {\begin{cases} \sin^{2} 2 x & n = 0 \\ 2^{2 n - 1} \sin (4 x + \frac{(n - 1) π}{2}) & n > 0 \end{cases}

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