Differentiate Sin2 x with respect to x …
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Posted by Innocent Geeta 7 years ago
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Naveen Sharma 7 years ago
Ans. By First Principle \(f'(x) = lim_{h\to 0} \space{ f(x+h) - f(x) \over h}\)
\(=> lim_{h\to 0} \space{ sin^2(x+h) - sin^2x \over h}\)
\(=> lim_{h\to 0} \space{ [sin(x+h) + sinx]\times [sin(x+h) - sinx]\over h}\) \([Using \space (a^2 - b^2) = (a+b)(a-b)]\)
\(=> lim_{h\to 0} \space{2 sin({x+h+x\over 2}) cos({x+h-x\over 2})\times 2cos({x+h+x\over 2}) sin({x+h-x\over 2})\over h}\)
\([Using \space sin a + sin b = 2 sin({a+b\over 2})cos ({a-b\over 2})\) and \( sin a - sin b = 2 cos({a+b\over 2})sin ({a-b\over 2})]\)
\(=> lim_{h\to 0} \space4{ sin({2x+h\over 2}) cos{h\over 2} \space cos({2x+h\over 2}) sin{h\over 2}\over h}\)
\(=> \space4 sin({2x+0\over 2}) cos{0\over 2} \space cos({2x+0\over 2}) lim_{h\to 0} {sin{h\over 2}\over {2h\over 2}}\)
\(=> 4 sin x. cos 0. cos x. lim_{{h\over 2}\to 0} {1\over 2}{sin{h\over 2}\over {h\over 2}}\) \([as \space h \to 0, then, {h\over 2}\to 0 ]\)
\(=> 4 sin x. cos x. {1\over 2}\) \([Using \space Identity, lim_{x \to 0 } \space {sinx \over x } = 1]\)
\(=> 2 sin x. cos x\)
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