Question

for f(x)=ln(e^(3x) + e^(-3x)) find f'(x) and evaluate at x=ln2

Answer 1

simply use chain rule

Answer 2

u know how to use it?

Answer 3

i'm not to sure can you guide me through?

Answer 4

take a basic example\[A=[f(g(x))]\]
now diffrentiating it with respect to x by chain rule--\[\frac{ dA }{ dx }=f'(g(x))*g'(x)\]

Answer 5

would it be \[\frac{ dA }{ dx }=\frac{ -9e ^{-9x} }{ e ^{3x}+e ^{-3x} }\]

Answer 6

no fisrt do a basic problem
\[A=\ln(x^2)\]
find the derivative

Answer 7

\[\frac{ 1 }{ x^2 }\]

Answer 8

yes here comes the problem ,here u have to use chain rule
\[\frac{ dA }{ dx }=\frac{ d }{ dx } \ln(x^2)=\frac{ d }{ dx }\ln(x^2)*\frac{ d }{ dx }(x^2)=\frac{ 1 }{ x^2 }*2x\]

Answer 9

getting this?

Answer 10

yep

Answer 11

* treat \( (e^{3x} + e^{-3x})\) as the \(x^2\) you just learned how to take the derivative of from @sidsiddhartha :)

Answer 12

yup can u do it now?

Answer 13

its like this right?
\[\frac{ 1 }{ e ^{-9x} }\times-9e ^{-9x}\]

Answer 14

When simplified, what does this give you?

Answer 15

\[\ \frac{d}{dx} (e^{3x}) = (e^{3x}) \cdot \frac{d}{dx} (3x)\] that is what I meant, sorry.

Answer 16

\[3e ^{3x}\]

Answer 17

Good.

Answer 18

And \[\ \frac{d}{dx} (e^{-3x}) = e^{-3x} \cdot \frac{d}{dx} (-3x)\]

Answer 19

What about this one? :)

Answer 20

\[-3e ^{-3x}\]

Answer 21

Awesome.

Answer 22

so answer to first part is \[\frac{ 3e ^{3x}-3e ^{-3x} }{ e ^{3x}+e ^{-3x} }\]

Answer 23

You're very very close.

Answer 24

I'll write up my understanding of it and you can try following my logic. Ok? :)

Answer 25

alright

Answer 26

So now that you've found the derivatives of both of those, we can find the derivative of the entire function. \[ \frac{d}{dx} \ln(e^{-3x} +e^{3x})\] By treating \(e^{-3x} + e^{3x}\) as \(x\) , we can think of this derivative as \(\frac{d}{dx} \ln(x)\) as \(\frac{1}{x}\).
We have found both of our derivatives for the inner function, which is good, but I think there is an easier way to go about solving this instead of finding the derivatives separately and then putting it all together.\[\ \frac{d}{dx} \ln(e^{-3x} +e^{3x})= \frac{1}{(e^{-3x} +e^{3x})} \cdot \frac{d}{dx}(e^{-3x} +e^{3x})\]Taking the derivative of the inside can be a little tricky, but what if we made the power's of e positive instead?

Answer 27

We would have \[\ \frac{1}{(e^{-3x}+e^{3x})}\cdot \left[-\frac{3}{e^{3x}}+3e^{3x}\right] \]What I did was I made \(e^{-3x}\) into a +ve fraction by setting it over 1, and then using the chain rule to take the derivative of the inner function, like what we did before, and i simply took the derivative of the other function as you also had done earlier.

Answer 28

Now focusing solely on the function in brackets, we need to write it so that we have a common denominator to combine the function altogether. we can multiply the top and bottom of \(3e^{3x}\) by \(e^{3x}\).\[\left[ -\frac{3}{e^{3x}} + \frac{3e^{3x}\cdot e^{3x}}{e^{3x}}\right] = \frac{-3 +3e^{3x+3x}}{e^{3x}}=\frac{-3+3e^{6x}}{e^{3x}}=\frac{3(-1+e^{6x})}{e^{3x}}\]

Answer 29

Following so far? :\

Answer 30

yep. i know it can be a very big hassle typing it out thanks for everything so far =)

Answer 31

Alright :) This solves the DERIVATIVE part of the function. Now we can concentrate on simplifying the major function as well. \[\frac{1}{(e^{-3x} +e^{3x})}\]rewriting this function, we can use the same method we used in finding the derivative. \[\large \frac{1}{\frac{1}{e^{3x}}+e^{3x}}=\frac{1}{\frac{1+e^{6x}}{e^{3x}}}\]And now we... combine EVERYTHING together. Oh god.

Answer 32

\[\large \frac{\frac{3(-1+e^{6x})}{e^{3x}(1+e^{6x})}}{e^{3x}}\]And cancelling the \(e^{3x}\) we get \[\frac{3(-1+e^{6x})}{(1+e^{6x})}\]

Answer 33

And that's how you solve this function :))!

Answer 34

oh god if that is one question on a previous midterm we had to do then i am so worried about my midterm. we have 5 pages of questions and that was only 1 question.

Answer 35

Oh, don't stress. Just take it one step at a time and review each step as you do it.

Answer 36

we have 60 mins for it =( a midterm i did before i ran out of time

Answer 37

Yeah... I feel that way when I take my calculus exams. you get used to working out hard problems really fast.

Answer 38

i feel so unprepared

Answer 39

Oh, don't feel that way! I suggest retrying this problem over again and seeing if you can solve it on your own.

Answer 40

Good luck! :)

Answer 41

thanks