What is the first derivative of $y= \sin^2(\sqrt{x})$? I'm trying to find out what the graph of the function $y= \sin^2(\sqrt{x})$ looks like. Can someone please help me?
 A: The graph of $ y = \sin^2(\sqrt{ x }) $ on $x \in [0,15]$ from MATLAB is:

The first derivative of $\displaystyle y = \sin^2(\sqrt{ x })$ is $\displaystyle y^{\prime} = 2 \sin (\sqrt{ x }) \cos( \sqrt{x} ) \frac{1}{2}x^{ - \frac{1}{2} } $.
Simplifying gives $\displaystyle y^{\prime} = \frac{ \sin{ \sqrt{x} } \cos{ \sqrt{ x } } }{ \sqrt{x} }$.
Which means the critical points are (when the numerator or denominator is 0):
$ \sin{ \sqrt{x} } = 0 $
Due to the graph of $\sin(x)$:

you know that's whenever the argument (value given) to $\sin$ is an integer multiple of $\pi$:
$\displaystyle \sqrt{x} = k \pi $
$\displaystyle x = k^2 \pi^2,~~ k \in {I} $
Also when $ \cos{ \sqrt{ x } } = 0 $, which is when

$ \sqrt{x} = k \dfrac{\pi}{2} $
$ x = \dfrac{ k^2 \pi^2 }{ 4 },~~ k \in {I} $
So you can see the first critical point is when k=1, and then we have:
From the $\cos$ equation:
$ x = \dfrac{ k^2 \pi^2 }{ 4 } $
$ x = \dfrac{ \pi^2 }{ 4 } \approx 2.47 $
Which you can see is the first peak in the graph,
From the $\sin$ equation:
$ x = k^2 \pi^2 $
$ x = \pi^2 \approx 9.87 $
Which is the second peak in the graph, at about $x=10$.
Here's a larger graph for $ k \in [ 0, 100 ] $:

A: Apply chain rule twice to get:
$\displaystyle y^{\prime} = 2 \sin(\sqrt{x}) \cos(\sqrt{x}) \frac{1}{2\sqrt{x}} = \frac{\sin(2 \sqrt{x})}{2\sqrt{x}}.$
Now note, that the critical points is where $~~ 2 \displaystyle\sqrt{x} = \pi k,~~ k = 1,2,3,\ldots$.
Find the zeros and note that they are precisely the minimal values.
