Find #(d^2y)/dx^2# if #y=e^sqrt(sin(x))#?

#y=e^sqrtsinx#

1 Answer
maganbhai P.
Jun 5, 2018

#y=e^sqrtsinx=>(dy)/(dx)=e^sqrtsinxd/(dx)(sqrtsinx)#
#=>(dy)/(dx)=e^sqrtsinx *1/(2sqrtsinx)d/(dx)(sinx)=e^sqrtsinx/(2sqrtsinx )xxcosx#
#=>(dy)/(dx)=(e^sqrtsinx *cosx)/(2sqrtsinx)#
Please see below.

Explanation:

Here,

#y=e^sqrt(sinx)#

Let,

#y=e^u,u=sqrtv andv=sinx#

#=>(dy)/(du)=e^u,(du)/(dv)=1/(2sqrtv) and(dv)/(dx)=cosx#

#"Using "color(blue)"Chain Rule : "#

#color(blue)((dy)/(dx)=(dy)/(du)xx(du)/(dv)xx(dv)/(dx)#

#(dy)/(dx)=e^uxx1/(2sqrtv)xxcosx#

Subst. #u=sqrtv#

#(dy)/(dx)=e^sqrtv xx1/(2sqrtv)xxcosx#

Subst. #v=sinx#

#(dy)/(dx)=e^sqrt(sinx)xx1/(2sqrt(sinx))cosx#

#=>(dy)/(dx)=(e^sqrtsinx *cosx)/(2sqrtsinx)#

#:.y_1=(y*cosx)/(2sqrtsinx), ...to(1)#

#where, y=e^sqrtsinx and (dy)/(dx)=y_1#

#=>2sqrtsinx*y_1=y*cosx#

Again diff.w.r.t.#x#

#2sqrtsinx*y_2+y_1*2/(2sqrtsinx)cosx=y_1*cosx+y*(-sinx)#

Using #(1)# we get

#2sqrtsinx*y_2+(ycosx)/(2sqrtsinx)cosx/(sqrtsinx)= (ycosx)/(2sqrtsinx)cosx-ysinx#

#=>2sqrtsinx*y_2+(ycos^2x)/(2sinx)=(ycos^2x)/(2sqrtsinx)-ysinx#

#=>2sqrtsinx*y_2=y[cos^2x/(2sqrtsinx)-cos^2x/(2sinx)-sinx]#

#=>2sqrtsinx*y_2=(y[sqrtsinxcos^2x-cos^2x-2sin^2x])/(2sinx)#

#=>y_2=(y[sqrtsinxcos^2x-cos^2x-2(1-cos^2x)])/(2sqrtsinx*2sinx)#

Thus,

#(d^2y)/(dx^2)=(e^sqrtsinx[sqrtsinxcos^2x+cos^2x- 2])/(4(sinx)^(3/2))#

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