find the derivative of xtanx /cosx+sinx
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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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