Find : `d/dx { e^( n ( log( a + x ) - log( a -x ) )) }` =

d/dx (e^(sin√x))

d/(dx) (e^(x sin x)) =

d/(dx)e^(log sqrt(x)}=

(d)/(dx)(log e^(x)*log x)

Differentiate the following with respect of x:e^(a log a)+e^(a log x)+e^(a log a)

(d)/(dx)[log((x)/(e^(tan x)))]=

(d)/(dx)(e^(x)+log x+tan x+x^(6))

The differentiation of log_(a)x(a>0,a)*!=1 with respect to x is (1)/(x log_(a)a) i.e.(d)/(dx)(log_(a)x)=(1)/(x log_(a)a)

The value of (d)/(dx)(log_(2)log_(3)x) at x=3 is (ln x=log_(e)x)

(d)/(dx){log_(e)(ax)^(x)}