Simplify...

[Imum Coeli]

Question:
Simplify:

i) d/dx e^2^x^2

Find:

ii) d/dx ln(((x+1)^4 * (4x-1)^4)/sqrt(x^2+3))

Notes:
i) I used a calculator to find the answer but have no idea how it got there.

ii) I think that
ln(((x+1)^4 * (4x-1)^4)/sqrt(x^2+3)) = ln((x+1)^4) + ln((4x-1)^4) - ln((x^2+3)^(1/2))
= 4*ln(x+1) + 4*ln(4x-1) - 1/2*ln(x^2+3)
so d/dx of the above is
4/(x+1) +16/(4x-1) + x/(x^2+3)
but that is different to the calculator's answer so somewhere I have made a mistake.

Thanks for any advice.

 

[stapel]

i) d/dx e^2^x^2

Is this any of the following?

\(\displaystyle \mbox{a) }\, \frac{d}{dx}\, e^{2^{\left(x^2\right)}}\)

\(\displaystyle \mbox{b) }\, \frac{d}{dx}\, e^{{\left(2^x\right)}^2}\)

\(\displaystyle \mbox{c) }\, \frac{d}{dx}\, {\left(e^2\right)}^{x^2}\)

\(\displaystyle \mbox{d) }\, \frac{d}{dx}\, {\left(e^{2^x}\right)}^2\)

Or something else?

ii) d/dx ln(((x+1)^4 * (4x-1)^4)/sqrt(x^2+3))

Is the above as follows?

\(\displaystyle \frac{d}{dx}\, \ln\left(\frac{(x\, +\, 1)^4\, \times \, (4x\, -\, 1)^4}{\sqrt{x^2\, +\, 3}}\right)\)

Thanks!

 

[Imum Coeli]

Sorry for not being clear.

From the writing I believe it is a)

The natural logarithm derivative is correct.

Sorry again.

 

[stapel]

I believe it is a)

So it's as follows:

. . . . .\(\displaystyle \mbox{a) }\, \frac{d}{dx}\, e^{2^{\left(x^2\right)}}\)

Note that "e^2" is just a number, so you've got a number raised to the power x^2. What rule would apply (in addition to the Chain Rule) for differentiating a number to a power containing a variable?

 

[Imum Coeli]

Thanks.

I think I have it.

d/dx (e^2^(x^2)) = d/du(e^u) = (dx/du)*(du/dx) where u=2^(x^2) and dx/du=e^u

d/dx (e^2^(x^2)) = e^u*d/dx(u) = (e^2^(x^2))*(d/dx(2^(x^2)))

(e^2^(x^2))*(d/dx(2^(x^2))) = (e^2^(x^2))*(d/dx(2^u)) where u=x^2 and dx/du=2^u*ln(2)

d/dx (e^2^(x^2)) = 2^u*ln(2)*(du/dx)*e^2^(x^2) = 2^(x^2)*ln(2)*(2x)*e^2^(x^2)

So 2^(x^2)*ln(2)*(2x)*e^2^(x^2) is the answer.

Thank you.

 

[lookagain]

Thanks.

I think I have it.

d/dx (e^2^x^2) = d/du(e^u) = (dx/du)*(du/dx) where u=2^x^2 and dx/du=e^u

d/dx (e^2^x^2) = e^u*d/dx(u) = (e^2^x^2)*(d/dx(2^x^2))

(e^2^x^2)*(d/dx(2^x^2)) = (e^2^x^2)*(d/dx(2^u)) where u=x^2 and dx/du=2^u*ln(2)

d/dx (e^2^x^2) = 2^u*ln(2)*(du/dx)*e^2^x^2 = 2^x^2*ln(2)*(2x)*e^2^x^2

So 2^x^2*ln(2)*(2x)*e^2^x^2 is the answer.

Thank you.

Imum Coeli, those lines of work above are not acceptable, because they include ambiguous expressions in the forms of a^b^c and a^b^c^d. \(\displaystyle \ \ \ \ \ \ \)They must have grouping symbols to eliminate ambiguity.

 

[Imum Coeli]

Sorry. I've tried to fix. I still may be missing a set of brackets. (I wish I knew how to use Latex properly!)
Sorry again.