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Grade: 12
        
differentiate the following function with respect to x
(cos x)^x + (sin x)^1/x
5 years ago

Answers : (2)

Rinkoo Gupta
askIITians Faculty
80 Points
							

Y1=(cosx)x

taking log on both sides

logy1=xlogcosx

diff. using product rule with respect to x

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

dy1/dx=(cosx)x[-xtanx+logcosx]

y2=(sinx)1/x

taking log on both sides

logy2=1/x logsinx

diff. using product rule with respect to x

(1/y2)dy2/dx=(1/x)(1/sinx)(cosx)+logsinx(-1/x2)

dy2/dx=(sinx)1/x[(cotx)/x-(1/x2)logsinx]

hence dy/dx=(cosx)x[-xtanx+logcosx]+(sinx)1/x[(cotx)/x-(1/x2)logsinx]

Thanks & Regards

Rinkoo Gupta

AskIITians Faculty

5 years ago
Arun Kumar
IIT Delhi
askIITians Faculty
256 Points
							
Hello
Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty
5 years ago
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