Differentiation of Exponential Functions

[Becky4paws]

I am doubting the way I did this problem...

Finding 1st and 2nd derivatives

f(x) = 2/(1+3e^-2x)

f(x) = (1+3e^-2x)^-1
f'(x) = (1+3e^-2x)(-1)(-2)
f'(x) = 2(1+3e^-2x)

f"(x) = (1+3e^-2x)(-2)
f'(x) = -2(1+3e^-2x)

 

[galactus]

Hey Becky:

You can use the product rule, but how about the quotient rule?.

\(\displaystyle \L\\f(x)=\frac{2}{1+3e^{-2x}}\)

\(\displaystyle \L\\f'(x)=\frac{(1+3e^{-2x})(0)-(2)(-6e^{-2x})}{(1+3e^{-2x})^{2}}\)

\(\displaystyle \L\\=\frac{12e^{2x}}{(e^{2x}+3)^{2}}\)

Differentiate this to get f''(x).

Now, run through it with the product rule and see if you arrive at the same thing.

 

[Becky4paws]

2nd differentiation

f'(x) = 12e^2x/(e^2x+3)^2

f"(x) = (e^2x+3)^2*(24e^2x) - (12e^2x)(2)(2x)/(e^2x+3)^4
f"(x) = (24e^2x) - (12e^2x)(-4x)/(e^2x+3)^2
f"(x) = 24e^2x-12e^2x(-4x)/(e^2x+3)^2
f"(x) = 12e^2x(-4x)/(e^2x+3)^2

This gets so confusing

 

[galactus]

\(\displaystyle \L\\f'(x)=\frac{12e^{2x}}{(e^{2x}+3)^{2}}\)

Quotient rule:

\(\displaystyle \L\\\frac{(e^{2x}+3)^{2}(24e^{2x})-(12e^{2x})4e^{2x}(e^{2x}+3)}{(e^{2x}+3)^{4}}\)

Now, one of the worst things of all this is not the calc, but the algebra.

\(\displaystyle \L\\\frac{-24e^{2x}(e^{2x}-3)(e^{2x}+3)}{(e^{2x}+3)^{4}}\)

\(\displaystyle \L\\=\frac{-24e^{2x}(e^{2x}-3)}{(e^{2x}+3)^{3}}\)

 

[soroban]

Re: 2nd differentiation

Hello, Becky4paws!

\(\displaystyle \L f'(x) \:= \:12\cdot\frac{e^{2x}}{(e^{2x}\,+\,3)^2}\)

\(\displaystyle \L f''(x) \;=\;12\,\cdot\,\frac{(e^{2x}\,+\,3)^2\,\cdot\,e^{2x}\cdot2 \:- \: e^{2x}\,\cdot\,2(e^{2x}\,+\,3)\cdot e^{2x}\cdot2}{(e^{2x}\,+\,3)^4}\)-

. . . . . . .\(\displaystyle \L=\;12\,\cdot\,\frac{2e^{2x}(e^{2x}\,+\,3)^2\:-\:4e^{4x}(e^{2x}\,+\,3)}{(e^{2x}\,+\,3)^4}\)

Factor: \(\displaystyle \L\:12\,\cdot\,2e^{2x}(e^{2x}\,+\,3)\,\cdot\, \frac{(e^{2x}\,+\,3)\,-\,2e^{2x}}{(e^{2x}\,+\,3)^4}\)

Simplify: \(\displaystyle \L\:24e^{2x}\cdot\frac{e^{2x}\,+\,3\,-\,2e^{2x}}{(e^{2x}\,+\,3)^3} \;=\;\frac{24e^{2x}\left(3\,-\,e^{2x}\right)}{(e^{2x}\,+\,3)^3}\)