How do we differentiate #sqrt(2sin(sqrtx))#?

1 Answer
Jan 8, 2018

#d/(dx)sqrt(2sin(sqrtx))=cos(sqrtx)/(2sqrt(2xsin(sqrtx))#

Explanation:

We use the concept of function of a function.

As #f(x)=sqrt(g(x))#, where #g(x)=2sin(h(x))# and #h(x)=sqrtx#

Hence #(df)/(dx)=(df)/(dg(x))xx(dg(x))/(dh(x))xx(dh)/(dx)#

Hence #d/(dx)sqrt(2sin(sqrtx))#

= #1/(2sqrt(2sin(sqrtx)))xx2cos(sqrtx)xx1/(2sqrtx)#

= #cos(sqrtx)/(2sqrt(2xsin(sqrtx))#