# 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?

-
Do you know the chain rule? Have you plotted the function? – Rasmus Mar 20 '11 at 12:42
@Rasmus, I've just looked up the chain rule but I'm not sure how to apply it. And yes I've plotted the function on excel, but I need to know the mathematical boundaries of it. – Chloe Mar 20 '11 at 12:46
Boundaries? Do you mean where the function is defined? $\sin$ takes all real arguments, and the square root function is defined for all non-zero $x$, so the domain is $[0, \infty)$. – Uticensis Mar 20 '11 at 13:07
adding to Billare, $\sin(x)$ gives values in $[-1,1]$, so $\sin^2(x)$ gives values in $[0,1]$ and so does $\sin^2(\sqrt{x})$. – Henry Mar 20 '11 at 16:57

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 $k \in [ 0, 100 ]$:

-
Very nice detailed answer! Also, I would suggest to write \in for the $\in$ symbol instead of \epsilon. – JavaMan Jun 3 '11 at 11:01

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.

-
Shouldn't your answer be divided by $2\sqrt{x}$ instead of $2x$? Also Chloe, note that shamovic used a trig. identity to change $2\sin (\sqrt{x}) \cos(\sqrt{x})$ into $\sin(2\sqrt{x})$. – Brian Mar 20 '11 at 13:32
Right, thanks. Edited, That is what happens when you don't pay attention to what you write :) – shamovic Mar 20 '11 at 13:37
@Brian, which trigonometric identity was that? – Chloe Mar 20 '11 at 13:49
@Shamovic, why are the critical points where $2 \sqrt{x} = \pi k$, $k = 1,2,3,\ldots$ ? I could only find that y'=0 when y'=\frac{\sin(2 \sqrt{x})}{2\sqrt{x}} implying that x=0 – Chloe Mar 20 '11 at 14:08
Well if a fraction is $0$, than means that it's nominator is $0$. Now you must think when does the sine function obtain it's zeros. – shamovic Mar 20 '11 at 14:18