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Maclaurin series for f(x) = sin 4x^2 is given by sin 4x = 4x^4/3! – 4x^6/5! + 4x^8/7! – 4x^10/9! + …
The Maclaurin series for cos 4x can be derived from the above series by taking the derivative of sin 4x and finding the derivative of both sides.
cos 4x = 1 – 4x^2/2! + 4x^4/4! – 4x^6/6! + …
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Hello students, given that Maclaurin’s series for fx equal to sine 4x is sine 4x equal to 4x minus 32 by 3 x cube plus 128 by 15 x to the power 5 minus 1024 by 315 x to the power 7 plus it goes to infinity. Now we differentiate both side. So differentiating we have 4 cos 4x equal to 4 minus 32 by 3 into 3 x square plus 128 by 15 into 5 x to the power 4 minus 1024 by 315 into 7 x to the power 6 up to infinity. So this is equal to this is equal to 4 minus 32 x…
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