How do you derive y = (1 + sin(2x)]/[1 - sin(2x)) using the quotient rule?

1 Answer
Topscooter
Jan 4, 2016

f'(x) = (4cos(2x))/(1-sin(2x))^2 = dy/dx.

Explanation:

By the quotient rule, the derivative of (u(x))/(v(x)) is (u'(x)v(x) - u(x)v'(x))/(v(x)^2). Let's say f(x) = (1+sin(2x))/(1-sin(2x)).

Here, u(x) = 1 + sin(2x) and v(x) = 1 - sin(2x). The derivative of sin(2x) is 2cos(2x) by the chain rule, so u'(x) = 2cos(2x) and v'(x) = -u'(x).

So, by applying the quotient rule :
f'(x) = (2cos(2x)(1-sin(2x)) - (1+sin(2x))(-2cos(2x)))/(1-sin(2x))^2

f'(x) = (2cos(2x)(1-sin(2x) + 1 + sin(2x)))/(1-sin(2x))^2 = (4cos(2x))/(1-sin(2x))^2

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