Derivatives of Exponential Functions

[f1player]

I'm having a little trouble with the following questions

Differentiate:

1. (e^x + e^-x)(x^2 +x+1)

2. x^2 e^(x^2) - 2xe^(x^2)

3. (e^x + e^-x) / (e^x - e^-x)

For number 1 I used the product rule and as an answer got:

(e^x + e^-x) (2x+1) + (x^2 +x +1)(e^x - e^-x)

Maybe could simplify it further???

The answer in the book states: e^x(x^2 +3x+2) - xe^-x(x-1)

???, that got me totally confused, how on earth did they get that???

Also for question 3, the answer was: (-4) / (e^x - e^-x)^2

How did they get the -4 there ???

Those 2 questions are the ones that confuse me the most
Any help would be appreciated

 

[galactus]

One way to arrive at number 3 is to use the quotient rule and the hyperbolic identities.

\(\displaystyle e^{x}+e^{-x}=2cosh(x)\), \(\displaystyle \frac{d}{dx}[2cosh(x)]=2sinh(x)\)

\(\displaystyle e^{x}-e^{-x}=2sinh(x)\), \(\displaystyle \frac{d}{dx}[2sinh(x)]=2cosh(x)\)

Using the quotient rule we get:

\(\displaystyle 4sinh^{2}(x)-4cosh^{2}(x)\)

\(\displaystyle sinh^{2}(x)-cosh^{2}(x)=-1\)

\(\displaystyle =4(-1)=-4\)

Now, all over the denominator squared.

The way you're probably looking for is most likely:

Use the quotient rule again.

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

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

=\(\displaystyle \L\frac{-4}{(e^{x}-e^{-x})^{2}\)

 

[soroban]

Hello, f1player!

Differentiate:

\(\displaystyle 1.\;\;(e^x\,+\,e^{-x})(x^2\,+\,x\,+\,1)\)

I used the product rule and as an answer got:
\(\displaystyle \;\;\;(e^x\,+\,e^{-x}) (2x\,+\,1) \,+\,(x^2\,+\,x\,+\,1)(e^x\,-\,e^{-x})\;\) . . . correct!

The answer in the book states: \(\displaystyle \,e^x(x^2\,+\,3x\,+\,2)\,-\,xe^{-x}(x-1)\)

How on earth did they get that? \(\displaystyle \;\) by doing a lot of simplifying ... with no warning.

\(\displaystyle \text{We have: }\L\,(e^x\,+\,e^{-x})(2x\,+\,1)\,+\,(x^2\,+\,x\,+\,1)(e^x\,-\,e^{-x})\)

\(\displaystyle \text{Expand: }\L\:2xe^x\,+\,e^x\,+\,2xe^{-x}\,+\,e^{-x}\,+\,x^2e^x\,-\,x^2e^{-x}\,+\,xe^x\,-\,xe^{-x}\,+\,e^x\,-\,e^{-x}\)

\(\displaystyle \text{Simplify: }\L\:x^2e^x\,+\,3xe^x\,+\,2e^x\,-\,x^2e^{-x}\,+\,xe^{-x}\)

\(\displaystyle \text{Factor: }\L\:e^x(x^2\,+_\,3x\,+\,2)\,-\,xe^{-x}(x\,-\,1)\)

\(\displaystyle 2.\;\; x^2 e^{x^2}\,-\,2xe^{x^2}\)

Use the Product Rule twice . . .

\(\displaystyle \L\;\;x^2\cdot e^{x^2}\cdot2x\,+\,2x\cdot e^{x^2}\:-\:2xe^{x^2}\cdot2x\,-\,2e^{x^2}\)

\(\displaystyle \L\;\;\;\;= \;2x^3e^{x^2}\,-\,4x^2e^{x^2}\,+\,2xe^{x^2}\,-\,2xe^{x^2}\)

\(\displaystyle \text{Factor: }\L\:2e^{x^2}(x^3\,-\,2x^2\,+\,x\,-\,1)\)

\(\displaystyle 3.\;\;\frac{e^x\,+\,e^{-x}}{e^x\,-\,e^{-x}}\)

The answer was: \(\displaystyle \,\frac{-4}{(e^x\,-\,e^{-x})^2}\)

How did they get the -4 there ? \(\displaystyle \;\) . . . more simplifying

\(\displaystyle \text{Quotient Rule: }\L\:\frac{(e^x\,-\,e^{-x})(e^x\,-\,e^{-x})\,-\,(e^x\,+\,e^{-x})(e^x\,+\,e^{-x})}{(e^x\,-\,e^{-x})^2}\)

\(\displaystyle \L\;\;=\;\frac{(e^{2x}\,-\,e^x\cdot e^{-x}\,-\,e^{-x}\cdot e^x\,+\,e^{-2x})\,-\,(e^{2x}\,+\,e^x\cdot e^{-x}\,+\,e^{-x}\cdot e^x\,+\,e^{-2x})}{(e^x\,-\,e^{-x})^2}\)

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

\(\displaystyle \L\;\;=\;\frac{-4}{(e^x\,-\,e^{-x})^2}\)

 

[f1player]

Yes, i understand now


Although the question only asked to find the derivative, not to simplify it to its simplest form.

So i was right after all...

Thanks for your help