Differentiate the function #y=log_e(cos2x)#. What is the area of the region enclosed by the curve #f (x)= tan(2x)#, the x axis and the lines #x=0 and x=\pi/8#?

1 Answer
Noah G
Dec 4, 2016

#dy/dx = (-2sin(2x))/cos(2x)#

Explanation:

I will answer the question about differentiating #y = log_e(cos2x)#.

First of all, #y= log_e(cos2x)# is equivalent to #y = ln(cos2x)#. Instead of using the chain rule to differentiate this twice, I would recommend changing this into exponential form.

#e^y = cos2x#

By the chain rule and implicit differentiation, we can differentiate. I'll start by showing you how to use the chain rule for #cos2x#.

We let #y = cosu# and #u = 2x#, then #dy/(du) = -sinu# and #(du)/dx= 2#.

#dy/dx= 2 xx -sinu#

#dy/dx = -2sin(2x)#

The entire function, now:

#e^y(dy/dx) = -2sin(2x)#

#dy/dx= (-2sin(2x))/e^y#

#dy/dx = (-2sin(2x))/(e^(ln(cos2x)))#

#dy/dx = (-2sin(2x))/cos(2x)#

Hopefully this helps!

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