d/dx(2^3^x^2)

[ahorn]

Question: differentiate \(\displaystyle 2^{3^{x^2}}\)

I found 4 ways of solving this problem, depending on what powers I took out of the natural log. Here I've written out 2 of them. The second one was correct according to the memo given to me. What have I done wrong in the first solution? Thank you



\(\displaystyle 2^{3^{x^2}}=e^{2\ln{2^{3^x}}}\\

\dfrac{d}{dx}e^{2\ln{2^{3^x}}}=e^{2\ln{2^{3^x}}} \cdot2 \cdot\frac{\frac{d}{dx}2^{3^x}}{2^{3^x}}\\

=e^{2\ln{2^{3^x}}} \cdot2\cdot\frac{2^{3^x}\cdot\ln{2^3}}{2^{3^x}}\\

=2^{3^{x^2}}\cdot2\cdot3\ln2\\

=2^{3^{x^2}}\cdot6\ln2\)



OR



\(\displaystyle 2^{3^{x^2}}=e^{3^{x^2}\ln2}\\

\frac{d}{dx}e^{3^{x^2}\ln2}=e^{3^{x^2}\ln2}\cdot \frac{d}{dx}3^{x^2}\ln2\\

=2^{3^{x^2}}\cdot \frac{d}{dx}e^{x^2\ln3}\cdot\ln2\\

=2^{3^{x^2}}\cdot e^{x^2\ln3}\cdot2x\cdot\ln3\cdot\ln2\\

=2^{3^{x^2}}\cdot 3^{x^2}\cdot2x\cdot\ln3\cdot\ln2\\\)

 

[Deleted member 4993]

Question: differentiate \(\displaystyle 2^{3^{x^2}}\)

I found 4 ways of solving this problem, depending on what powers I took out of the natural log. Here I've written out 2 of them. The second one was correct according to the memo given to me. What have I done wrong in the first solution? Thank you



\(\displaystyle 2^{3^{x^2}}=e^{2\ln{2^{3^x}}}\\\) ............Incorrect

\(\displaystyle 2^{3^{x^2}}=e^{[\ln{2^{3^{x^2}}}]} \\ = e^{[{3^{x^2}}*\ln(2)]}\\\)

\(\displaystyle \dfrac{d}{dx}e^{2\ln{2^{3^x}}}=e^{2\ln{2^{3^x}}} \cdot2 \cdot\frac{\frac{d}{dx}2^{3^x}}{2^{3^x}}\\

=e^{2\ln{2^{3^x}}} \cdot2\cdot\frac{2^{3^x}\cdot\ln{2^3}}{2^{3^x}}\\

=2^{3^{x^2}}\cdot2\cdot3\ln2\\

=2^{3^{x^2}}\cdot6\ln2\)



OR



\(\displaystyle 2^{3^{x^2}}=e^{3^{x^2}\ln2}\\

\frac{d}{dx}e^{3^{x^2}\ln2}=e^{3^{x^2}\ln2}\cdot \frac{d}{dx}3^{x^2}\ln2\\

=2^{3^{x^2}}\cdot \frac{d}{dx}e^{x^2\ln3}\cdot\ln2\\

=2^{3^{x^2}}\cdot e^{x^2\ln3}\cdot2x\cdot\ln3\cdot\ln2\\

=2^{3^{x^2}}\cdot 3^{x^2}\cdot2x\cdot\ln3\cdot\ln2\\\)

.

 

[HallsofIvy]

Your first step would be correct if the formula were \(\displaystyle \left(2^{3^x}\right)^2= 2^{2(3^x)}\) but that is NOT the same as \(\displaystyle 2^{3^{x^2}}\)

 

[ahorn]

Oh, thank you. I see now that I was taking the wrong term to ^2. I should only take x to ^2 and not 2^(3^x) because that would be saying 2^(2(3^x)).