Question:

Differentiate w.r.t. xx the function: sin3x+cos6xsin^3x+cos^6x

Updated On: Oct 19, 2023
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Solution and Explanation

The correct answer is =3sinxcosx(sinx2cos4x)=3sinxcosx(sinx-2cos^4x)
Let y=sin3x+cos6xy=sin^3x+cos^6x
dydx=ddx(sin3x)+ddx(cos6x)∴\frac{dy}{dx}=\frac{d}{dx}(sin^3x)+\frac{d}{dx}(cos^6x)
=3sin2x.ddx(sinx)+6cos5x.ddx(cosx)=3sin^2x.\frac{d}{dx}(sinx)+6cos^5x.\frac{d}{dx}(cosx)
=3sin2x.cosx+6cos5x.(sinx)=3sin^2x.cosx+6cos^5x.(-sinx)
=3sinxcosx(sinx2cos4x)=3sinxcosx(sinx-2cos^4x)
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