I have some derivatives questions???HELP!!!(10 POINTS FOR BEST ANSWER, NO JOKE)?

Question

Hey, I have these two questions:

Find the equation of the normal line of y = 2^(2x) / ln(x) at x=1

I understand the question and what its asking for and I also know how to do it, but when I try to find the y-point, the bottom becomes 0 because the ln(1) = 0, so I dont know what to do if the corresponding y-value is undefined.

Here is the second question:

Find all points on the unit circle where the slope of the tangent line of y = tan(x) at x = pi/4

I find the derivative which is (sec(x))^2 and I find the zeroes of that but it comes out to be no solution or undefined. I don't know how to do the problem from there.

PLEASE PLEASE SHOW ALL STEPS, AND PLEASE WRITE EVERYTHING DOWN, DON'T SKIP OR BE VAGUE.

I PROMISE 10 POINTS WILL BE AWARDED TO THE BEST ANSWERER.

Answer 1

You are correct that the function is undefined at x=1.



There can be no normal line "at x=1".

There's a vertical asymptote whose equation is x=1, but there cannot be a normal line to discontinuity, especially a nonremovable discontinuity such as this one. The left- and right-hand limits are not equal (one goes to -inf, and the other to +inf)









The second question has bizarre wording.





Either you're looking for points on the unit circle, or you're looking for points on the graph y = tan x.



The slope of the line tangent to the unit circle at angle θ is tan θ, so it's possible they're just looking for all the places where the tangent line has the same slope as the tangent line at θ.



If that's the case, you're looking for the two places that have a slope equal to tan(pi/4), which is -1. These places would be pi/4 and 5pi/4.

Answer 2

The derivative doesn't exist at x=1 because there is an asymptote there. I would just write "The derivative does not exist at the given x value."

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sec²(x) doesn't equal zero at any point. I'm not totally sure what this question is asking, but it sounds as if it wants all the points between [0, 2π] where the slope of tan(x) is the same as the slope of tan(π/4). If that's the case then find the slope at π/4 and then solve for all the x values that produce that slope.



y′(π/4) = sec²(π/4) = 2



2 = sec²(x)

x = arcsec(±√2)

x = π/4, 3π/4, 5π/4, 7π/4

Answer 3

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Answer 4

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