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| 1. |
Differentiate Sin2\xa0x with respect to x from 1st principle method\xa0 |
| Answer» Ans. By First Principle \\(f\'(x) = lim_{h\\to 0} \\space{ f(x+h) - f(x) \\over h}\\)\\(=> lim_{h\\to 0} \\space{ sin^2(x+h) - sin^2x \\over h}\\)\\(=> lim_{h\\to 0} \\space{ [sin(x+h) + sinx]\\times [sin(x+h) - sinx]\\over h}\\)\\([Using \\space (a^2 - b^2) = (a+b)(a-b)]\\)\\(=> lim_{h\\to 0} \\space{2 sin({x+h+x\\over 2}) cos({x+h-x\\over 2})\\times 2cos({x+h+x\\over 2}) sin({x+h-x\\over 2})\\over h}\\)\\([Using \\space sin a + sin b = 2 sin({a+b\\over 2})cos ({a-b\\over 2})\\)\xa0and\xa0\\( sin a - sin b = 2 cos({a+b\\over 2})sin ({a-b\\over 2})]\\)\\(=> lim_{h\\to 0} \\space4{ sin({2x+h\\over 2}) cos{h\\over 2} \\space cos({2x+h\\over 2}) sin{h\\over 2}\\over h}\\)\\(=> \\space4 sin({2x+0\\over 2}) cos{0\\over 2} \\space cos({2x+0\\over 2}) lim_{h\\to 0} {sin{h\\over 2}\\over {2h\\over 2}}\\)\\(=> 4 sin x. cos 0. cos x. lim_{{h\\over 2}\\to 0} {1\\over 2}{sin{h\\over 2}\\over {h\\over 2}}\\)\\([as \\space h \\to 0, then, {h\\over 2}\\to 0 ]\\)\\(=> 4 sin x. cos x. {1\\over 2}\\)\\([Using \\space Identity, lim_{x \\to 0 } \\space {sinx \\over x } = 1]\\)\\(=> 2 sin x. cos x\\) | |