How do you differentiate #g(x) = [1 + sin(2x)]/[1 - sin(2x)] # using the product rule?

1 Answer
Guilherme N.
Jan 9, 2016

In order to forcedly use the product rule, we'll rewrite it as #g(x)=(1+sin(2x))(1-sin(2x))^-1#

Explanation:

Also, we'll need chain rule to differentiate the second term.

  • Chain rule: #(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)#

  • For the first term, we'll rename #u=2x#

  • For the second term, we'll rename #v=(1-sin(w))# and #w=2x# (but the logic follows the same way that of #u#, #v#, and so on)

  • Product rule: #(ab)'=a'b+ab'#

#(dg(x))/(dx)=2cos(2x)(1-sin(2x))^-1+(1+sin(2x))-v^-2(-2cos(2x))#

Substituting #v# and rearranging negative exponents:

#(dg(x))/(dx)=(2cos(2x))/(1-sin(2x))+(2cos(2x)(1+sin(2x)))/(1-sin(2x))^2#

#(dg(x))/(dx)=(2cos(2x)(1-sin(2x))+2cos(2x)(1+sin(2x)))/(1-sin(2x))^2#

#(dg(x))/(dx)=(2cos(2x)(1cancel(-sin(2x))+1+cancel(sin(2x))))/(1-sin(2x))^2#

#(dg(x))/(dx)=(4cos(2x))/(1-sin(2x))^2#

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