Thread: To differentiate a trig func. using the logarithms, need your opin's on my solution

Sorry for not using LaTeX, I'm not used to it yet, but I promise I would improve, so the equation is:

Let f(x)= 6tan^(-1)(4e^(x). Need to find the derivative

I continued to solve the equation, before differentiating f(x) = y = 6(1/(tan(4e^(x))) = 6cot (4e^(x)),

Now, ln y = ln (6cot (4e^(x))) =>
ln y = ln 6 + ln(cot(4e^(x))) =>

y’/y = (1/6) + (1/ cot(4e^(x))) *(- csc^(2) (4e^(x)))*(4xe^(x-1))) =>

y’/y = (1/6) - [(4xe^(x-1))*(csc^(2)(4e^(x)]/[cot(4e^(x)], and =>

y’ = 6cot (4e^(x)) * {(1/6) - [(4xe^(x-1))*(csc^(2)(4e^(x))]/[(cot(4e^(x))]},

I definitely have some problems with log functions, I think...

Thank you

Originally Posted by dokrbb

Sorry for not using LaTeX, I'm not used to it yet, but I promise I would improve, so the equation is:

Let f(x)= 6tan^(-1)(4e^(x). Need to find the derivative

Is it possible that really is the inverse tangent

do you mean that I had to do in this way:

f(x) = y = 6(arctan(4e^(x))),

Now, ln y = ln (6arctan (4e^(x))) =>
ln y = ln 6 + ln(arctan(4e^(x))) =>

y’/y = (1/6) + (1/ (1 + (4e^(x))^(2)) * (4xe^(x-1)))

y’/y = (1/6) + (4xe^(x-1))/(1 + (16e^(2x)), and

y’ = 6(arctan(4e^(x))* {(1/6) - [(4xe^(x-1))]/[(1 + (16e^(2x))]},

(I wonder, do I have to rewrite and solve for ln y = ln 6 + x ln(arctan(4e), ot that doesn't change the result )

y’/y = (1/6) + ln(arctan(4e) + x*[1/ (1 + (4e^(x))^(2))] * (4e) =>


y’ = 6(arctan(4e^(x))* {(1/6) + ln(arctan(4e) + [(4xe)]/[(1 + (16e^(2x))]},


well. I just tried to solve by two ways, and now I'm not sure whether both of them is wrong or only the first one...

Originally Posted by dokrbb

do you mean that I had to do in this way:
f(x) = y = 6(arctan(4e^(x))),

If it is indeed , why in the world are you doing anything with logarithms?

this would sound as a dumb answer, but it's a homework set of problems after discussing the "Logarithmic differentiation", the teacher's argument was that sometimes, when we have some complex functions, would be easier (sounds funny after comparing your answer with mine ) to use log differentiation; Like I said - is a dumb answer, but I assumed they want us to apply the Log differentiation for this chapter ...

Originally Posted by dokrbb

this would sound as a dumb answer, but it's a homework set of problems after discussing the "Logarithmic differentiation", the teacher's argument was that sometimes, when we have some complex functions, would be easier (sounds funny after comparing your answer with mine ) to use log differentiation; Like I said - is a dumb answer, but I assumed they want us to apply the Log differentiation for this chapter ...

Quite often textbooks will throw in a question that has nothing to do with the current topic. It is a way of review and keeps students on their toes.