Question

Logarithmic differentiation of f' when y=(Inx)^cosx (x>1)

Answer 1

`(dy)/(dx) = (lnx)^(cosx ) \cdot [(cosx )/(ln x) \cdot (1)/(x) - sin x ( ln ( ln x)) ]`

Explanation:

`y = (ln x)^(cosx)\\ln y = cos x (ln ( ln x)) \\`
`[ taking \ logarithm \ on \ both \ sides ]`

`(1)/(y) (dy )/(dx) = (cosx )/(ln x) \cdot (1)/(x) + ln ( ln x) (-sin x)`
`[Chain rule : uv = u (dv)/(dx) + v (du)/(dv) ]`

`(dy)/(dx) = y * [(cosx )/(ln x) \cdot (1)/(x) - sin x ( ln ( ln x)) ]`

`(dy)/(dx) = (lnx)^(cosx ) \cdot [(cosx )/(ln x) \cdot (1)/(x) - sin x ( ln ( ln x)) ]`
`[ substituting \ the \ value \ of \ y]`