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

Question

Isaiah Allen · Accepted Answer

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

Chain rule will also be required in order to distinguish 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#

Ezekiel Alexander · Answer

undefined To differentiate $ g(x) = \frac{1 + \sin(2x)}{1 - \sin(2x)} $ using the product rule, first, identify the two functions as $ u(x) = 1 + \sin(2x) $ and $ v(x) = 1 - \sin(2x) $. Then, apply the product rule, which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

The derivative of $ u(x) $ is $ u'(x) = 2\cos(2x) $, and the derivative of $ v(x) $ is $ v'(x) = -2\cos(2x) $.

Now, apply the product rule formula:

$$ g'(x) = u'(x)v(x) + u(x)v'(x) $$

Substitute the derivatives and the original functions into the formula:

$$ g'(x) = (2\cos(2x))(1 - \sin(2x)) + (1 + \sin(2x))(-2\cos(2x)) $$

Simplify the expression:

$$ g'(x) = 2\cos(2x) - 2\sin(2x)\cos(2x) - 2\cos(2x) - 2\sin(2x)\cos(2x) $$

Combine like terms:

$$ g'(x) = 2\cos(2x) - 2\cos(2x) - 2\sin(2x)\cos(2x) - 2\sin(2x)\cos(2x) $$

$$ g'(x) = -4\sin(2x)\cos(2x) $$

So, the derivative of $ g(x) $ with respect to $ x $ using the product rule is $ -4\sin(2x)\cos(2x) $.