y=x^sinx
log y= log x^sinx
log y= sinx logx
1/y(dy/dx) = cosx logx + (sinx)/x
dy/dx= y[cosx logx = 1]
dy/dx = x^sinx (cosx log x + 1)
Written on Sept. 10, 2019, 9:02 a.m.
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Answer of this question
Differentiate x^sinx, x>0 w.r.t. x.
Shruti Shibu
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