Question

What is the derivative of `6 (3^(2x-1))`?

The answer following the book is `4 ln(3)3^(2x)`. I think it is wrong.
I think the answer should be `12(3^(2x-1))ln3`

Answer 1

Both are same. `(dy)/(dx)=4*ln3*3^(2x)` or `12*3^(2x-1)*ln3`

Explanation:

For questioner

In fact both are same as

`12*3^(2x-1)*ln3`

= `4*3^1*3^(2x-1)*ln3`

= `4*3^(1+2x-1)*ln3`

= `4*ln3*3^(2x)` - observe that this answer is less complicated.

For others , who may like to know

As `y=6*3^(2x-1)=2*3*3^(2x-1)=2*3^(1+2x-1)=2*3^(2x)`

Now taking log on both side (base `e`), we get

`lny=ln2+2xln3`

and differentiating `1/y*(dy)/(dx)=2*ln3`

or `(dy)/(dx)=2*ln3*y`

= `2*ln3*(2*3^(2x))`

= `4*ln3*3^(2x)`

Answer 2

The answer given by the book and your answer are equivalent. I shall show this by starting with your answer:

`12(3^(2x-1))ln(3)`

Because addition (or subtraction) of exponents is equal to the multiplication of the base with the two exponents, I can separate `3^(2x-1)` into `3^(2x)3^-1`:

`12(3^(2x))(3^-1)ln(3)`

Because a negative exponent is the same as division, I can write `3^-1` as `1/3`:

`12(3^(2x))1/3ln(3)`

We finish by observing that `12/3 = 4`:

`4(3^(2x))ln(3)`

This is the same as the answer given by the book.