What is the first derivative of $y= \sin^2(\sqrt{x})$?

Question

I'm trying to find out what the graph of the function $y= \sin^2(\sqrt{x})$ looks like. Can someone please help me?

Answer 1

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 ] $:

Answer 2

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.