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find the derivative of \((\cos { x } +i\sin { x) } (\cos { 2x+i\sin { 2x)(\cos { 3x } +i\sin { 3x) } ......(cosnx+i\sin { nx) } } } \)

Note by Rishabh Jain 3 years, 6 months ago

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2 \times 3

2^{34}

a_{i-1}

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e^ix(1+2+3+....+n)=e^ixn(n+1)/2 So derivative is in(n+1)/2e^ixn(n+1)/2

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6*e^i((π/2)+6x)

Use De Moivre's formula, \({ e }^{ inx }=cos(nx)+isin(nx)\)

just tell me the answer i know it's easy.

\({ e }^{ i\cdot [n(n+1)/2]\cdot x }\cdot i\cdot n(n+1)/2\)

@Akash Shah – correct answer

use de moivre's theorem an then it will come like e power n into n+1 into x into i then use chain rule and differentiate

such a simple one but i cant type the answer properly

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`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNeweste^ix(1+2+3+....+n)=e^ixn(n+1)/2 So derivative is in(n+1)/2e^ixn(n+1)/2

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6*e^i((π/2)+6x)

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Use De Moivre's formula, \({ e }^{ inx }=cos(nx)+isin(nx)\)

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just tell me the answer i know it's easy.

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\({ e }^{ i\cdot [n(n+1)/2]\cdot x }\cdot i\cdot n(n+1)/2\)

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use de moivre's theorem an then it will come like e power n into n+1 into x into i then use chain rule and differentiate

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such a simple one but i cant type the answer properly

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