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Find the derivative of xtanx /cosx+sinx Plz if u know the answer then answer otherwise no need |
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Answer» Let y=xtanx/(sinx+cosx). After rearranging, ysinx + ycosx = xtanx Differentiating both SIDES and USING multiplication rule i.e (xy)'=xy'+yx' y' sinx+ ycosx +y'cosx - ysinx = TANX + sec^2(x) After doing SIMPLE algebraic rearrangment we will get: y'= (tanx + xsec^2(x) - xtanx.cos2x)/(sinx+cosx) Hope this helps.Step-by-step explanation:Note: (cosx - sinx)/(cosx + sinx) = cos2x |
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