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Math Help - trigonomic derivatives problem

  1. #1
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    Question 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!!
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  2. #2
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    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)
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  3. #3
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    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)
    Thanks from NJoyce
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