Question

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

Answer 1

\[\large d(3^{2x^2-3x})\]This follows the format: \(d(a^x) = a^x \cdot d(x)\cdot \ln(a)\)

Answer 2

So let's first find the derivative of the power. Can you solve : \(d(2x^2-3x)\)?

Answer 3

@dolphins121 ?

Answer 4

uhmm

Answer 5

the derivative of that would be 4x-3 right?

Answer 6

Yes, then just use the form to find your answer :)

Answer 7

So what would \(a^x\) be? and \(\ln(a)\)?

Answer 8

a^x would be 3^4x-3 ?

Answer 9

That would be \(a^{d(x)}\)

Answer 10

so a^x is the original problem?

Answer 11

yep :)

Answer 12

and ln(a) is 3?

Answer 13

nope. \(\ln(a) = \ln(3)\)

Answer 14

because \(a=3\)

Answer 15

yeah my bad thats what i meant. so the answer is 3^(4x-3) x ln(3) ?

Answer 16

nope, fit the format.

\(\color{blue}{\text{Originally Posted by}}\) @Jhannybean
\[\large d(3^{2x^2-3x})\]This follows the format: \(d(a^x) = a^x \cdot d(x)\cdot \ln(a)\)
\(\color{blue}{\text{End of Quote}}\)

Answer 17

ohh so it's 3^(2x^2-3x) ⋅ 3^(4x-3) ⋅ ln(3) ?

Answer 18

Let me refresh, it's showing question marks :(

Answer 19

oh should i rewrite it?

Answer 20

Oh! you got it :) Good job. I got that too :D

Answer 21

thanks to you ! :)

Answer 22

sorry about the notation as well, \(d(3^{2x^2-3x}) = \frac{d}{dx}(3^{2x^2-3x}) = (4x-3)3^{2x^2-3x}\cdot \ln(3)\)

Answer 23

its np. thank you so much !

Answer 24

Good job :)

Answer 25

thanks :)