Question 1

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

Question 2

Answer:

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

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) ]

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

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

Please log in or register to add a comment.

Please log in or register to answer this question.