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Home Q & A Board Science-Related Homework Help Mathematics
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tony tony
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11 years ago
How would I go about using the quotient rule (I know what it is) to solve the following:

g(x) = sin(x)/cos(x)

I understanding that the dervatives of sin(x) and cos(x) are cos(x) and -sin(x), respectively.
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wrote...
#1
11 years ago
The quotient rule is

f(x) = g(x)/h(x)
f'(a) = [ g'(a)h(a) - g(a)h'(a) ] / h(a)^2
therefore in your question

g(x) = sin(x)
g'(x) = cos(x)
h(x) = cos(x)
h'(x) = -sin(x)

therefore f'(x) = [ cos(x)cos(x) + sin(x)sin(x) ] / cos(x) ^2
f'(x) =[ cos(x)^2 + sin(x)^2 ] cos(x)^2

since cos(x)^2 + sin(x)^2 = 1
you get
f'(x) = 1/cos(x)^2
wrote...
#2
11 years ago
Well, you COULD say that g(x) = tan(x), then just take the derivative of tan(x) = sec^2(x)

If you are bound and determined, however, remember the quotient rule says:
g(x)f'(x)-f(x)g'(x)/ g(x)^2
where f(x) is the top and g(x) is the bottom.  Using our problem then, we have:
(cosx)(cosx)-(sinx)(-sinx) / (cosx)^2

The top will reduce to cosx^2+sinx^2, which of course reduces to 1!  So, we get 1 / (cosx)^2  which is also secx^2.
Answer accepted by topic starter
tony1069tony1069
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11 years ago
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