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#?

Answer 1

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

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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