# Math Help - trigonomic derivatives problem

1. ## trigonomic derivatives problem

Hey everyone,
So I've been trying for days to find the first positive maximum of the function
(Cos x) * ( sqrt ( 20 sin x ) )
(Ie. highest point on the curve)
I can't even differentiate it, but even if I do figure out the gradient formula and let it equal zero, I don't understand how to solve for x with all those sinea and cosines. I know from looking at the graph that the answer is close to 35, but I need to prove it mathematically. Any help would be greatly appreciated. :)

Ps. I need explanations of how to do it, not just an answer please!

Thankyou!!

2. ## Re: trigonomic derivatives problem

First use the product rule,
$\text{let }u=cos(x) \text{ and }v=\sqrt{20sin(x)}$

When differentiating v use the chain rule,
$\text{let }w=20sin(x)$

3. ## Re: trigonomic derivatives problem

Hello, NJoyce!

Find the first positive maximum of the function: . $f(x) \:=\: \cos x \sqrt{20\sin x }$

I would simplify first . . .

$f(x) \;=\;\cos x\cdot\sqrt{20}\cdot\sqrt{\sin x} \;=\;\sqrt{20}\cdot\cos x\cdot (\sin x)^{\frac{1}{2}}$

$f'(x) \;=\;\sqrt{20}\left[ \cos x\cdot\tfrac{1}{2}(\sin x)^{-\frac{1}{2}}\cos x + (-\sin x)(\sin x)^{\frac{1}{2}}\right] \;=\;0$

. . . . . . $\sqrt{20}\left[\frac{\cos^2\!x}{2(\sin x)^{\frac{1}{2}}} - (\sin x)^{\frac{3}{2}}\right] \;=\;0$

. . . . . . $\sqrt{20}\left[\frac{\cos^2\!x-2\sin^2\!x}{2(\sin x)^{\frac{1}{2}}}\right] \;=\;0$

Then: . $\cos^2\!x - 2\sin^2\!x \:=\:0 \quad\Rightarrow\quad \2\sin^2\!x \:=\:\cos^2\!x$

. . . . . . $\frac{\sin^2\!x}{\cos^2\!x} \:=\:\frac{1}{2} \quad\Rightarrow\quad \tan^2\!x \:=\:\frac{1}{2} \quad\Rightarrow\quad \tan x \:=\:\pm\frac{1}{\sqrt{2}}$

Therefore: . $x \;=\;\arctan\left(\pm\frac{1}{\sqrt{2}}\right)$