How do you find the derivative of #sin^2(sqrtx)#?

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Answer 1 · 2017-03-17T20:57:22

#(dy)/(dx)=(sin(sqrtx)cos(sqrtx))/(sqrtx)#

Let #y=u^2#, where #u=sin(sqrtx)#

Since we have a function within a function, we must use the chain rule to derive #y#, where #(dy)/(dx)=(dy)/(du)*(du)/(dx)#

#d/(du)u^2=2u=2sin(sqrtx)#

To derive #u# in terms of #x#, we need to apply chain rule once more.

Let #u=sin(v)#, #v=sqrtx#

#(du)/(dv)=cos(v)#, #(dv)/(dx)=1/2x^(-1/2)=1/(2sqrtx)#

Hence, #(du)/(dx)=(du)/(dv)*(dv)/(dx)=cos(sqrtx)*1/(2sqrtx)#

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

Now that we know #(du)/(dx)#, we can find #(dy)/(dx)#

#(dy)/(dx)=(dy)/(du)*(du)/(dx)=(cancel(2)sin(sqrtx)cos(sqrtx))/(cancel(2)sqrtx)#

#=(sin(sqrtx)cos(sqrtx))/(sqrtx)#