Question

Question #bbdc7

Answer 1

`= 6e^(2x)`

Explanation:

We will require the use of the chain rule: If `u` is a function of `x` and `y` is a function of `u`, then:

`(dy)/(dx) = (dy)/(du) (du)/(dx)`

If we let `u = 2x`, then:

`d/dx(3e^(2x)) = d/(du)(3e^u) * d/dx(u)`

`=3e^u * 2 = 6e^u = 6e^(2x)`. (since `d/dx(2x) = 2`)

Note:
If `y` is a function of `x`, then `d/dx(cy) = c(dy)/(dx)`, where `c` is a real, nonzero constant.

Answer 2

`6e^(2x)`

Explanation:

the standard derivative of `e^x" is " e^x`

To differentiate `e^(2x)" use the" color(blue)" chain rule"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(d/dx(e^(f(x)))=e^(f(x)) .f'(x))color(white)(2/2)|)))`

`rArrd/dx(3e^(2x))=3e^(2x).d/dx(2x)=6e^(2x)`

Answer 3

see explanation.

Explanation:

`color(orange)"Reminder"`

`ln(xy)=lnx+lny`

`rArrln(3e^(2x))≠2xln(3e)`

`"but "ln(3e^(2x))=ln3+lne^(2x)=ln3+2xcancel(lne)`

`"and "d/dx(ln3+2x)=2`

`rArrm'(x)=2m(x)=2.3e^(2x)=6e^(2x)`