Question

What is the derivative of sqrt(3^x)

Answer 1

\[y = \sqrt{3^{x}}\]

yes?

Answer 2

Yes

Answer 3

long term aim is expressing this using the natural number, e, right?? cos e was literally *made* for differentiation and integration.

start by getting rid of the root sign. ideas?

Answer 4

So, 1/2(3^x)^-1/2?

Answer 5

Rewrite the root into fractional exponent, then follow chain rule

Answer 6

sadly not!!!

this is an exponential, and this is where you need the natural number.

\[\frac{d (e^{x})}{dx} = e^{x}\]

only works for e, and that is how e is calculated.

you need ultimately to shift 3^x into something that has an e in it

Answer 7

@nincompoop is a **qualified helper** so i will let nincompoop take the lead.....

Answer 8

And a few other rules

Answer 9

So I believe with the chain rule I would have to multiply by ln(3), correct?

Answer 10

I am on an iPad so don'twait for my responses

Answer 11

The website is spotty and laggy for me

Answer 12

Okay

Answer 13

Use u-substitution

Answer 14

u = 3^x

Answer 15

Okay, I'm not sure if I did it right, but using U-substitution I got ln3/(2*3^x)^1/2

Answer 16

Show us how you end up with that

Answer 17

1/2(u)^-1/2 which is 1/2(u)^1/2 and then I replaced u with 3^x and applied the chain rule multiplying ln3* 1/2*(3^x)^1/2

Answer 18

Do you know how to perform "implicit" differentiation?

Answer 19

I'm not sure I remeber

Answer 20

e.g. would you know how to differentiate this?\[y^2=3x\]

Answer 21

would it be the square root of 3?

Answer 22

using "implicit" method

Answer 23

I'm not sure, that was my best guess at it

Answer 24

what methods have you been taught?

Answer 25

Standard and u-substitution

Answer 26

Try following this video to learn about implicit method: https://www.khanacademy.org/math/differential-calculus/taking-derivatives/implicit_differentiation/v/implicit-differentiation-1

it will help you to answer your question - let me know if you still require help after watching this and I will try and explain the method more

Answer 27

@debins33
you must know that:

\[\sqrt{3^{x}} = 3^{\frac{x}{2}}\]

now, the rest should be easy for you.

the substitution suggested by the "qualified helper" was unhelpful. being pointed toward Youtube by someone else must count as equally unhelpful.

take this and what you have already shown you know and this is easy.

Answer 28

@IrishBoy123 - I would like to see how you approach this - please continue...

Answer 29

The Khan Academy video helped until the end, I didn't understand the final solution, but I'm not sure I completely know how to apply it to this problem

Answer 30

In this particular problem we can proceed as follows:\[\begin{aligned}
y&=\sqrt{3^x}\\
\therefore y^2&=3^x\\
\end{aligned}\]Next we would take the natural log of both sides to obtain:\[\begin{aligned}
\ln(y^2)&=\ln(3^x)\\
\therefore 2\ln(y)&=x\ln(3)\end{aligned}\]We can now apply the implicit method

Answer 31

So xln(3)/2ln

Answer 32

That is not how the method works - it looks like you have just tried to divide both sides of the equation I got to by \(2\ln\) - this is incorrect.

Please ping me when you are back online and I will try to explain to you how to use the implicit method.

Answer 33

@asnaseer

QED....

this thing can be done in so many ways, yet the point of this website is not to mislead and confuse.

if you can't see the simplest way of doing this, you are learning/ have learned calculus by rote and are applying it as such.

the opening proposition can be simplified into terms that the OP clearly understands. in simple algebra.

Answer 34

Which is why I asked you to proceed - but you seemed to decline

Answer 35

QED again.

can you seriously not differentiate an exponential without resorting to implicit differentiation?!?!

that was my suspicion. that is why i made the point in the first place.

you really ought to be able to decompose something as simple as root (3^x) and then switch it into a power of e before you bandy about the results of implicit diff.

do you actually know where e comes from?

you challenge is to turn 3 ^ x into something that can be differentiated without resorting to stuff that the OP clearly has not yet learned. do that or prove my point for the 3rd time!

lol.

Answer 36

@IrishBoy123 - I will let you believe what you want to believe. I am here to try and help others - not to argue.

Answer 37

we can all use a sledge hammer to crack a nut.

so do this problem by turning 3 ^ x into something that can be differentiated.

it is actually so simple.