Can someone help me with deriving using the quotient rule?

Question

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.

Answer 1

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 2

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

Answer 3

Quotient rule states
f(x) = f1(x)/f2(x)
f'(x) = [ f2(x)*f1'(x) - f1(x)*f2'(x) ] / [f2(x)] ^ 2
so substituting above
g'(x) = [ cos(x) * cos (x) - sin (x) * -sin(x) ] / [cos (x)] ^2
And cos2(x)+sin2(x) = 1
g'(x) = 1/[cos(x)]^2 or sec(x) ^2 if you prefer