 # find dy/dx wheny=x^(cos x) + (sin x)^ tan x

9 years ago

Let y1=xcosx

taking log on boths ides, we get

logy1=cosx.logx

now diff.w.r.to x , we get

(1/y1)dy1/dx=cosx.(1/x)+(logx).(-sinx)

dy1/dx=y1[(cosx)/x-sinx.(logx)]

=xcosx[(cosx)/x –sinx .logx]

let y2=sinxtanx

taking log on both sides, we get

logy2=tanxlogsinx

diff. w.r. to x , get

(1/y2)dy2/dx=tanx.(1/sinx).cosx+logsinx.sec2x

dy2/dx=sinxtanx[1+sec2x . logsinx]

Hence dy/dx = xcosx[(cosx)/x –sinx .logx]+ sinxtanx[1+sec2x . logsinx]

Thanks & Regards

Rinkoo Gupta Kushagra Madhukar
3 years ago
Dear student,

Let u = xcosx
taking log on boths ides, we get
logu = cosx.logx
now diff.w.r.to x , we get
(1/u)du/dx = cosx.(1/x) + (logx).(–sinx)
du/dx = u[(cosx)/x – sinx.(logx)]
= xcosx[(cosx)/x – sinx.logx]

let v = sinxtanx
taking log on both sides, we get
logv = tanx.log(sinx)
diff. w.r. to x , get
(1/v)dv/dx = tanx.(1/sinx).cosx + logsinx.sec2x
dv/dx = sinxtanx[1 + sec2x.logsinx]

Hence,
dy/dx = du/dx + dv/dx
= xcosx[(cosx)/x – sinx .logx] + sinxtanx[1 + sec2x.logsinx]

Thanks and regards,
Kushagra