Question

find the derivative of (3^2x)/x

Answer 1

(u/v)' = (vu' - u v') / v^2

Answer 2

how do i set it up? find u and du?

Answer 3

u=3^2x
v=x

Answer 4

That is the general rule for taking the derivative of a quotient.
\( \dfrac{d}{dx} \left( \dfrac{3^{2x}}{x} \right) = \dfrac{d}{dx} \left( \dfrac{u}{v} \right) \)
Just compare the numerator and denominator. u = 3^2x and v = x. You see that better here?

Answer 5

ah, so is it u sub or what do i do next

Answer 6

You'll need to compute the derivative of u and v.
Then its a direct substitution into that rule formula.

Answer 7

so du is 0 and dv=1???

Answer 8

Woah, careful. \( 3^{2x} \) is not a constant. Do you know the derivative of exponential functions?

Answer 9

\[xdy/dx(3^2^{x}) - 3^2^{x}dy/dx(x)/x^2\]

Answer 10

@pnp97

Answer 11

In a sense of exponential functions with a non-e base, you can make the adjustment with a logarithm property:
\( N = e^{\ln N} \)
\( 3^{2x} = \left(3^x\right)^2 = \left( e^{x \ \ln 3}\right)^2 \)
\( \ln 3\) is a constant now. And the derivative of e^u is just itself. multiplied by the derivative of u.

Answer 12

oh so once you adjust it to ln3 the derivative of that is itself so du=ln3

Answer 13

@rvc is that it solved completely???

Answer 14

\( \dfrac{d}{dx} \left( 3^{2x} \right) = \dfrac{d}{dx} \left( e^{2 \ln 3 x} \right) = e^{2 \ln 3 x} * 2 \ln 3 \)
Note chain rule comes in to play.
This one is understandably tricky. So take your time and think about it. :)

Answer 15

\( \dfrac{d}{dx} \left( 3^{2x} \right) = \dfrac{d}{dx} \left( e^{2 \ln 3 \ \ x} \right) = e^{2 \ln 3 \ \ x} * 2 \ln 3 \)
That x is not inside the argument of \( \ln 3\).

Answer 16

After that, you are correct in the d/dx ( x) = 1
So we would then just directly substitute those expressions we found:
\( \dfrac{d}{dx} \left( \dfrac{u}{v} \right) =\dfrac{du/dx \ \ v - u \ \ dv/dx}{v^2} \)

Answer 17

alright so du is e^2ln3x*2ln3????

Answer 18

Yes. (a small simplification is also to reverse that trick with the logarithm to make it look more like the u= 3^(2x) ==> \(du/dx = e^{2 \ln 3 \\ x} * 2 \ln 3 = 3^{2x} * 2 \ln 3\)

Answer 19

ad do after that it would be d/dx (3^2x)/(x)) = (3^2x*2ln3)/dx (whats dx?) x-3^2x 1/dx)/1

Answer 20

\( \dfrac{d}{dx} \left( \dfrac{3^{2x}}{x} \right) = \dfrac{3^{2x} 2 \ln 3 \times x - 1 \times 3^{2x}}{x^2} \)
I should emphasize that it is du/dx is the whole expression of the derivative of u with respect to x. du is just a differential and is equal to the derivative times dx.

Answer 21

Despite the variables being u and v (like an integration by parts), du/dx and dv/dx are the full expressions of derivatives. du/dx = u'(x). du = u'(x) dx.

Answer 22

that makes sense! does the expression have to be simplied and combine the like terms i.e. 3^2x??/

Answer 23

for calculus, i know my teacher was lenient about it. as long as you had all the derivatives simplified out, you could leave it at that point. it might just depend on what your teacher considers though.

Answer 24

i think it should be fine! thanks so much!

Answer 25

You're welcome! :)

Answer 26

All the best @pnp97