first take ln of both sides
ln y = ln(x^-cosx)
ln y = -cosx lnx
now differentiate with respect to x
(dy/dx)/y = -cosx (1/x) + lnx (sinx)
dy/dx = y ( -(1/x)cosx + lnx(sinx))
= x^-cosx ( -(1/x)cosx + lnx(sinx))
How do you find the logarithmic differentiation of y = x^-cos x ?
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