differentiate y=x^cosx with respect to x.?

Answer 1

y = x^(cosx)

taking logs

ln(y) = cosx ln(x)

differentiating

(1/y)dy = [cos(x)(1/x) + ln(x)(-sin(x)]dx

dy/dx = y[(1/x)cos(x) - sin(x) ln(x)]

dy/dx = x^(cosx)[(1/x)cos(x) - sin(x) ln(x)]

Answer 2

implicit differenciation

y = x^(cosx)

take nateral log for both sides
lny = ln x^(cosx)

log power rule:
ln(a^x) = x ln(a)

lny = cosx ln(x)

dy (lny) = d/dx [(cosx) ln(x)]

dy 1/y = d/dx (cosx) ln(x)

product rule:
d/dx (uv) = u'v + v'u

dy 1/y = d/dx (cosx) lnx + d/dx (lnx) (cosx)

dy 1/y = (-sinx)lnx + (cosx)/x

dy/dx = y * [(-sinx)lnx + (cosx)/x]

dy/dx = y [(cosx)/x - sinx lnx)

recall that y = x^(cosx)

dy/dx = x^(cosx) [(cosx)/x - sinx lnx] <== answer


Rec

Answer 3

Apply product rule of differentiation:
d(uv) = du x v + u x d(v)
If u = x^, then d(x^) = n.x power (n-1) .
If v = Cos x, then d(Cos x) = -Sin x.
As such, d(x^ cos x) = n.x power (n-1) xCos x - x^ Sinx.