Given: #f(x)=sin(sqrtarcsinx)#
We'll use the chain rule but to make this easier to visualize we are going to let #sqrt(arcsinx)=u#
So we have
#f(x)=sin(u)#
By the chain rule, we are solving
#f'(x)=d/dx[sin(sqrtarcsinx)]=color(red)((df)/(du)[sin(u)])*color(blue)((du)/dx[sqrtarcsinx]#
The derivative of #color(red)(sin(u)# is #color(red)(cos(u))#
#:.color(red)((df)/dx[sin(u)]=cos(u)#
For #color(blue)((du)/dx[sqrtarcsinx]# we can use the chain rule again.
We can call #sqrtarcsinx# a new function, let's call it #g(x)#
#g(x)=sqrtarcsinx=>color(blue)(g'(x)=(dg)/dx[sqrtarcsinx]=(du)/dx[sqrtarcsinx]#
And similarly we'll let #arcsinx=v#
So we have
#g(x)=sqrtv#
By the chain rule we are differentiating
#color(blue)((dg)/dx[sqrtarcsinx])=color(purple)((dg)/(dv)[sqrtv]*color(green)((dv)/dx[arcsinx]#
For #color(purple)((dg)/(dv)[sqrtv]=>d/dx[v^(1/2)]#
By the Power Rule
#color(purple)((dg)/(dv)=1/2v^(1/2-1)=1/2v^(-1/2)=1/(2v^(1/2))=1/(2sqrtv)#
#:.color(purple)((dg)/(dv)[sqrtv]=1/(2sqrtv)#
For #color(green)((dv)/dx[arcsinx]#, it's derivative is #color(green)(1/sqrt(1-x^2)#
#:.color(green)((dv)/dx[arcsinx]=1/sqrt(1-x^2)#
Now that we know what #color(purple)((dg)/(dv) and color(green)((dv)/dx# are, we can find #color(blue)((dg)/dx[sqrtarcsinx])#
#color(blue)((dg)/dx[sqrtarcsinx])=color(purple)((dg)/(dv)[sqrtv])*color(green)((dv)/dx[arcsinx])=1/(2sqrtv)*1/sqrt(1-x^2)=1/(2sqrtvsqrt(1-x^2))#
Okay, now let us summarize
#u=sqrtarcsinx " " v=arcsinx#
#color(red)((df)/dx[sin(u)]=cos(u) " " color(blue)((du)/dx[sqrtarcsinx]=1/(2sqrtvsqrt(1-x^2)))#
Going back to the original problem:
#f'(x)=color(red)((df)/(du)[sin(u)])*color(blue)((du)/dx[sqrtarcsinx])=color(red)(cos(u))*color(blue)(1/(2sqrtvsqrt(1-x^2)))#
Now we have to substitute in back the values for #u# and #v# which will give us the final result:
#f'(x)=color(red)(cos(sqrtarcsinx))*color(blue)(1/(2sqrtarcsinxsqrt(1-x^2)))#
#f'(x)=color(red)(cos(sqrtarcsinx))/color(blue)((2sqrtarcsinxsqrt(1-x^2)))#
I hope this helped! :)
