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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