Math, asked by Khansanaya439, 4 months ago

find the derivative of xtanx /cosx+sinx


Plz if u know the answer then answer otherwise no need ​

Answers

Answered by subhsamavartj
0

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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