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 $(dy)/(dx)$

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

$=(sin(sqrtx)cos(sqrtx))/(sqrtx)$

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