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December 31st, 2014, 05:44 PM   #1
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Differentiation

I'm given $\displaystyle y=\frac{cos2x}{1-sinx}$ to differentiate.

My answer is $\displaystyle \frac{cosxcos2x-2(sin2x)(1-sinx)}{(1-sinx)^2}$

But the given answer is $\displaystyle -\frac{4sin(2x)-3cos(x)+cos(3x)}{2(sin(x)-1)^2}$.
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December 31st, 2014, 11:04 PM   #2
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Post your work and we'll help you identify and correct any errors.
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January 1st, 2015, 12:10 AM   #3
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$\displaystyle y=\frac{cos2x}{1-sinx}$

$\displaystyle \frac{dy}{dx}=\frac{(1-sinx)(-2sin2x)-(cos2x)(-cosx)}{(1-sinx)^2}$

$\displaystyle =\frac{-2sin2x+2sin^22x+cosx(cos2x)}{(1-sinx)^2}$

$\displaystyle =\frac{(-2)(sin2x)\left [ 1-sinx \right ]+cosx(cos2x)}{(1-sinx)^2}$

$\displaystyle =\frac{cosxcos2x-2(sin2x)(1-sinx)}{(1-sinx)^2}$

The given answer is $\displaystyle -\frac{4sin(2x)-3cos(x)+cos(3x)}{2(sin(x)-1)^2}$
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January 1st, 2015, 09:43 AM   #4
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The given answer is correct. Your mistake is in assuming

$\displaystyle (-\sin(x))(-2\sin(2x))=2\sin^2(2x)$

Correct that, then use the identities

$\displaystyle \cos(\theta)-\cos(\varphi)= -2\sin\left(\frac{\theta+\varphi}{2}\right)\sin \left(\frac{\theta-\varphi}{2}\right)$

and

$\displaystyle \cos(\theta)+\cos(\varphi)=2\cos\left(\frac{\theta +\varphi}{2}\right)\cos\left(\frac{\theta-\varphi}{2}\right)$

to simplify further.
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January 3rd, 2015, 08:31 AM   #5
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What this symbol $\displaystyle \varphi$ means?
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January 3rd, 2015, 09:31 AM   #6
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It's a variable, phi, like theta is a variable.
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