Question

how to find the derivative of 3^(2^(x)^2-3x)

Answer 1

\[3^{2^{x^2}-3x}\]
this?

Answer 2

yes

Answer 3

\[y=\frac{3^{2^{x^2}}}{3^{3x}} \text{ First I'm going to rewrite this using law of exponents }\]
Now I will take ln( ) of both sides :

\[\ln(y)=\ln(3^{2^{x^2}})-\ln(3^{3x})\]
Now I'm going to use power rule for log...
\[ln(y)=2^{x^2} \ln(3)-3x \ln(3)\]
Now differentiate both sodes...
We may find that we have to find the derivative of 2^(x^2) on some scratch paper by itself.

Answer 4

ok how do you do that?

Answer 5

using ln( ) on both sides then differentiating both sides

Answer 6

how do you differentiate on both sides?

Answer 7

\[u=2^{x^2} \\ \ln(u)=x^2 \ln(2) \text{ now differentiate }\]

Answer 8

well you should know the derivative of x^2 is 2x
and the derivative of ln(u) is u'/u

Answer 9

well yeah but why didn't u take the derivative of 2x^2

Answer 10

its not 2 to the power of x squared. It's 2x^2

Answer 11

So are you telling me the problem I posted above was not the correct interpretation of the problem you posted?

Answer 12

yeah I'm sorry. That was my fault. i didn't realize you put 2 to the power of x^2. it's just supposed to be 2x^2

Answer 13

\[3^{2x^2-3x} ?\]

Answer 14

yes that is correct

Answer 15

\[y=3^{2x^2-3x} \\ \text{ still you need to take } \ln( ) \text{ of both sides } \\ \ln(y)=(2x^2-3x)\ln(3)\]
now just differentiate both sides
(this problem is a lot easier than the one before since we have less powers)

Answer 16

\[(\ln(y))'=(2x^2-3x)'\ln(3)\]

Answer 17

okay so next i take the derivative of (2x^2-3x) ?

Answer 18

well you take the derivative of both sides

Answer 19

the derivative of ln(y) =?
the derivative of (2x^2-3x)=?

Answer 20

uhm so 4x-3..

Answer 21

\[(\ln(y))'=(2x^2-3x)'\ln(3) \\ (\ln(y))'=(4x-3)\ln(3)\]
still need to differentiate the left hand side

Answer 22

uhm im not quite sure how to do that..

Answer 23

derivative (w.r.t. x) of ln(x) is 1/x
derivative of ln(u) (w.r.t x) is u'/u
so here you have y's instead of u's

Answer 24

are you still here

Answer 25

derivative of ln(y) (w.r.t x ) is just y'/y

Answer 26

\[\frac{y'}{y}=(4x-3) \ln(3)\]

Answer 27

solve for y'

Answer 28

and then replace y with what you started with (since that is what we called y)

Answer 29

okay so the answer is y'/y= (4x-3)ln(3) ?

Answer 30

well have you solve for y' yet?
and replace y with what we called y in the beginning
because that is what the lines above your post say to do...

Answer 31

look here is an example:
\[u=5^x \\ \ln(u)=\ln(5^x) \\ \ln(u)=x \ln(5) \\ \frac{u'}{u}=1 \ln(5) \\ \frac{u'}{u}=\ln(5) \\ u'=u \ln(5) \\ u' = 5^x \ln(5) \]

Answer 32

to solve for u' I multiplied both sides by u

Answer 33

u was called 5^x
so I replaced u with 5^x

Answer 34

here is another example
\[v=5^{x^2-3x+1} \\ \ln(v)=(x^2-3x+1)\ln(5) \\ \frac{v'}{v}=(2x-3)\ln(5) \\ v'=(2x-3) \ln(5) v \\ v'=(2x-3)\ln(5) 5^{x^2-3x+1}\]

Answer 35

so y'= 3^(2x^2-3x)(4x-3)(ln3) ?

Answer 36

sounds great

Answer 37

thank you so much! So i have another problem that I'm stuck on as well. Do you mind guiding me through that one? I understand if you're too busy but if you can i'll really appreciate it.

Answer 38

i have to eat
sorry

Answer 39

oh okay. well thanks anyways:)