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Grade: 12
        
find dy/dx when
y= x^x +(sin x)^ x
6 years ago

Answers : (1)

Rinkoo Gupta
askIITians Faculty
80 Points
							

Let y1=xx

Taking log on both sides, get

Logy1=xlogx

Diff w. r. to x ,get

1/y1. Dy1/dx=x.1/x+logx.1

dy1/dx=y1[1+logx]

=xx[1+logx]

Ley y2=(sinx)x

Logy2=xlogsinx

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

=xcotx+logsinx

dy2/dx=y2[cotx+logsinx]

=(sinx)x[cotx+logsinx]

Hence dy/dx = xx[1+logx]+ (sinx)x[cotx+logsinx]

Thanks & Regards

Rinkoo Gupta

AskIITians Faculty

6 years ago
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