Dreivatives and Logs

[EndersCatalyst]

I really need help on these. I have no idea how to do them.

Find the derivative:
1. f(x)= (4-e^-3x)^3
I did this. f'(x)= 3(4-e^-3x)^2 But I know it's not right.

2. f(x)= (e^-x)/ (1+x^2)
I got this far. f'(x)= [(1+t^2)e^x-e^x(2t)]/(1+t^2)^2

I also have no idea how to find the tangent line of y=e^-x^2 at (1,1/e). I know how, I just can't find the derivative. I got y'= (e^-2x)(-2)+(-2x)(e^-x^2)(-2x). I'm not sure how to combine the terms and simplify it, if it's even close to correct.

Last one, I swear. Find the derivative of f(x)= 2 ln x^5. I don't even know how to start this one.

 

[galactus]

Find the derivative:
1. f(x)= (4-e^-3x)^3
I did this. f'(x)= 3(4-e^-3x)^2 But I know it's not right.

Unfortunately, it is not correct. You did not implement the chain rule. What you have so far is good. You did not finish.

The chain rule is essentially the derivative of the inside times the derivative of the outside.

The derivative of the outside you have OK. The derivative of the inside is \(\displaystyle 3e^{3x}\)

So, we get \(\displaystyle 3(4-e^{-3x})^{2}\cdot 3e^{-3x}=9e^{-9x}(4e^{3x}-1)^{2}\)

2. f(x)= (e^-x)/ (1+x^2)
I got this far. f'(x)= [(1+t^2)e^x-e^x(2t)]/(1+t^2)^2

This one looks pretty good from what I can tell. Except, tidy it up a little. Why did you change from an x to a t?.

\(\displaystyle \frac{-(x^{2}+2x+1)e^{-x}}{(x^{2}+1)^{2}}\)

I also have no idea how to find the tangent line of y=e^-x^2 at (1,1/e). I know how, I just can't find the derivative. I got y'= (e^-2x)(-2)+(-2x)(e^-x^2)(-2x). I'm not sure how to combine the terms and simplify it, if it's even close to correct.

The derivative of \(\displaystyle e^{-x^{2}}\) is \(\displaystyle -2xe^{-x^{2}}\). Chain rule again.

Now, using the given point, the slope is \(\displaystyle m=\frac{-2}{e}\). Use this, along with the given point, and sub into y=mx+b. Solve for b and that's it.

Last one, I swear. Find the derivative of f(x)= 2 ln x^5. I don't even know how to start this one.

Is that\(\displaystyle f(x)=2ln(x^{5})\). Remember your log law. That is, \(\displaystyle ln(x^{n})=nln(x)\)

What is the derivative of ln?. That's about it.

 

[EndersCatalyst]

Thanks so much! I really needed the help.

One question though. How did you get 3e^3x as the derivative of 4-e^-3x?