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Differentiate Sin2 x with respect to x …

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Differentiate Sinx with respect to x from 1st principle method

 

  • 1 answers

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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