Question

how do you differentiate 4^3^x^2?

Answer 1

u bring the power behind the intiger then minus 1 ! for example if u have x^3 its gonna be 3x^2

Answer 2

that might work for x^2, but im pretty sure you use the chain rule to solve this problem, because its a constant to a constant to a variable to a constant

Answer 3

you are talikng about\[4^3\]to the \[x^4\]power, correct?

Answer 4

sorry\[ x^2\] power

Answer 5

Latex is failing to show that much to the power.

Answer 6

yes 4^(3^(x^(2)))

Answer 7

ok, notice that 4^3 is a constant
4^3=64
so we have
\[64^{x^2}\]and\[{d \over dx}(a^x)=a^x \ln( a)\]with that and the chain rule you should be able to solve this

Answer 8

Alright,it's not much difficult,just apply the chain rule repeatedly. Answer would be:

\[ 2^{1+6^{x^2}} \times 3^{x^2} \times x \times \text{ln}(3) \text{ln}(4) \]

Answer 9

when i use the chain rule should i make f(x)=x^2, then g(x)=64^x?

Answer 10

The way colton has written the function means we can't 'contract' the 4^3.

With y = 4^(3^(x^2))
then

ln y = 3^(x^2) ln 4
and thus

ln(ln y) = x^2 . ln 3 + ln(ln 4)

Now differentiate both sides, using implicit differentiation on the left hand side.

Answer 11

You may also check here: http://www.wolframalpha.com/input/?i=derivative+of+4^%283^%28x^%282%29%29%29

Answer 12

if it's g(f(x)) then yes

Answer 13

wow, I never anticipated such a difficult problem. I figure it had to be easier. that was clearly a mistake.
?

Answer 14

im just in the beginning of calc 1, so i dont know implicit differentiation yet, but i believe that i can do 4^3 first. when i wrote the equation i was trying to make it clearer

Answer 15

No James know what he's talking about and so does wolfram and colton, I was wrong, you cannot do that

Answer 16

check the link colton posted

Answer 17

foolformath's link has the equation written correctly

Answer 18

If the function is f(x) = 4^(3^(x^2)) and you can't use implicit differentiation, the chain rule will do it.

Let g(x) = x^2, h(x) = 3^x and j(x) = 4^x

then f(x) = j(h(g(x)))

and\[{df \ \over dx} = {dj \over dh}{dh \over dg}{dg \over dx}\]

Answer 19

ok thanks

Answer 20

This is a bit tricky. Try differentiating the function h(g(x)) first. And then layer on j