Q. 135.0( 1 Vote )

Let y=(logx)

Answer :

Given Data:


y=(logx)x+ xxcosx


Let u = (logx)x and v = xxcosx



Considering u = (logx)x


Taking log,


logu = x log(logx)


Differentiating w.r.t x,





Considering v = xxcosx


Taking log,


logv=x cosx logx


Differentiating w.r.t x,





Now y = u + v



Using the above values,



OR


Given Data:


x= a sin(pt), y=b cos(pt)


Differentiating w.r.t t




Now divide 2nd equation with 1st, we get



Again differentiating w.r.t x we get



Using the above value, we get




At t = 0, sec 0 =1



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