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Derivative - Quotient Rule Options
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benjamin.daniel12
#1 Posted : Friday, November 18, 2011 10:05:36 PM
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\frac{\mathrm{d} \frac{3(1-\sin x)}{2\cos x}}{\mathrm{d} x}
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Mattlink92
#2 Posted : Tuesday, November 22, 2011 8:10:57 PM
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Mattlink92
#3 Posted : Tuesday, November 22, 2011 8:41:26 PM
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The quotient rule is generally a nasty way to go about differentiating. It's really only convienient when the denominator is some square root.

\frac{\mathrm{d} }{\mathrm{d} x} \frac{3}{2}\left (\frac{1-sin(x)}{cos(x)} \right )

first move the constant outside the differentiate

\frac{3}{2}\frac{\mathrm{d} }{\mathrm{d} x}\left (\frac{1-sin(x)}{cos(x)} \right )

where

f(x) = 1-sin(x)

so f'(x) = -cos(x)

and

g(x) = cos(x)

so g'(x) = -sin(x)

 

using the quotient rule:

\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)} = \frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}

we have

\frac{\mathrm{d} }{\mathrm{d} x}\frac{1-sin(x)}{cos(x)} = \frac{-cos^2(x)+sin(x)-sin^2(x)}{cos^2(x)}

now simplify:

-1 + tan(x)sec(x)-tan^2(x)

then use trig identities to put it into any for convenient for you. dont forget to put your constants back on in front

\frac{3}{2}\left [-1 + tan(x)sec(x)-tan^2(x) \right ]
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